A student wants to know if the effect of gravity is the same on all objects. She follows these steps:

Drop a 5-gram ball, a 10-gram ball, and a 12-gram ball from a height of 10 meters.
Record how long it takes for each object to hit the ground.
?
Which step should be next?

(1 point)

Drop the balls from a different height.

Choose new balls.

Measure the mass of the balls again.

Perform a second trial.

The force F1 is equal and opposite to the downward force thus, F1 is equal to 60 N. The force F2 is inclined to 30 ° from leftward force and it is equal to 38.97 N in magnitude.

Force is an external agent acting on a body to deform it or to change its state of motion or rest. Force is a vector quantity and it is characterised by its magnitude and direction.

If two forces acting on a body from the same directions, then the net force will be the sum of these two forces. If they are acting from opposite directions, they will cancel each other in magnitude.

The force F1 is equal and opposite to the force acting downward. Thus its magnitude is 60 N. The force F2 is inclined to 30 ° from horizontal direction.

F2 = 45 cos 30 = 38.9 N.

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The work done by the external force on the child is positive.

When a force is applied to an object, work is done on that object. Work is defined as the product of the force applied on an object and the distance over which the force acts. In this case, the external force acted on the child on a skateboard, causing her speed to increase from 2 m/s to 3 m/s.

To calculate the work done, we can use the formula for work:

\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]

Since the child's speed increased, we know that the force and displacement acted in the same direction. Therefore, the angle between the force and displacement vectors, denoted by theta (θ), is 0 degrees, and the cosine of 0 degrees is 1.

Considering the child's speed increased, we can conclude that the force applied in the direction of motion did positive work on the child. The positive work done by the external force resulted in an increase in the child's kinetic energy, causing her speed to change.

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To determine the reaction at the beam supports for the given loading when ωo = 155 lb/ft, we have to follow some steps.

The steps are as follow:
Step 1: Identify the type of beam and support conditions. (e.g., simply supported, cantilever, overhanging, etc.)
Step 2: Determine the total length (L) of the beam.
Step 3: Calculate the total load (W) on the beam by multiplying the distributed load ωo by the length L: W = ωo * L.
Step 4: Identify the location and magnitude of any additional point loads, if applicable.
Step 5: Use equilibrium equations to find the reactions at the supports:
a) Sum of vertical forces = 0: R1 + R2 = W (total load)
b) Sum of moments about one of the supports = 0: M1 = R1 * L1 - W * L2
Step 6: Solve the equilibrium equations for the unknown reactions R1 and R2.
Once you have completed these steps, you will have determined the reaction at the beam supports for the given loading when ωo = 155 lb/ft.

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Answer:

the speed that the car is going at an instant of time is called the instantaneous speed.

Answer:Jane drives at an average speed of 45 mph on a journey of 135 miles.

Explanation:

Answer:

Explanation:

Remark

He's pulled upwards by the parachute. The formula is given below. Details are found   searching for terminal velocity.

\(V_t=\sqrt\frac{2mg}{\rho A C_d}\)

Whatever that calculation turns out to be, a person has to be going a whole lot slower before he can safely hit the ground. The parachute is designed so that the person will be slowed very quickly to a safe velocity so when he hits the ground the impulse will not be as great as it could be. The force that the parachute delivers acts upward and opposes the force of gravity which is a downward force.

Answer:

a = 2 m/s²

v₄ = 8 m/s

Explanation:

We can find the final speed at the end of eight second by using first equation of motion:

\(v_{f} = v_{i} + at\\\)

where,

vf = final velocity at 8th second = 16 m/s

vi = initial velocity = 0 m/s

a = acceleration = ?

t = time = 8 s

Therefore, using the values in the equation we get:

\(16\ m/s = 0\ m/s + a(8\ s)\\\\a = \frac{(16\ m/s)}{8\ s}\)

a = 2 m/s²

Now, we can apply same equation of motion for 4 seconds of motion to find the velocity at the end of 4th second (v₄):

\(v_{4} = 0\ m/s + (2\ m/s^2)(4\ s)\\\)

v₄ = 8 m/s

If she increases this force by 19%, her acceleration will be 0.19 times the acceleration due to gravity, resulting in an acceleration of 1.862 m/s^2.

To find the force required for the window washer to raise herself slowly at a constant speed, we need to consider her weight. The weight of an object can be calculated by multiplying its mass by the acceleration due to gravity.

In this case, the mass of the person plus the bucket is given as 61 kg. Therefore, the weight of the person plus the bucket is (61 kg) × (9.8 m/s^2) = 598.8 N.

Since the window washer wants to raise herself at a constant speed, the force she exerts downward must be equal to her weight. Therefore, she needs to pull downward with a force of 598.8 N.

If she increases this force by 19%, we can calculate her acceleration.

The increase in force is 19% of 598.8 N, which is

(19/100) × 598.8 N

= 113.772 N.

This additional force will contribute to her acceleration. Since acceleration is given by the equation F = ma, where F is the net force and m is the mass, we can rearrange the equation to solve for acceleration.

In this case, the net force is the sum of her weight force (598.8 N) and the additional force (113.772 N). The mass remains the same (61 kg).

Therefore, the net force is

(598.8 N + 113.772 N)

= 712.572 N.

Plugging the values into the equation,

we get 712.572 N = (61 kg) × a,

which gives us the acceleration

a = 712.572 N / 61 kg

= 1.862 m/s^2.

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Answer: Speed and velocity both involve distance traveled over time, but velocity also includes direction

Explanation: speed is the time rate at which the object travels & velocity is the time rate at which the object travels also but includes direction too.  

The force of the storm is 0.42 times typical maximum force of friction.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

If the air strikes a person at the rate of 40 kg/s per square meter.

The person  is 1.50 m high and 0.50 m wide.

Net force = 40 × 1.5 × 0.50 × 9.8 N = 294 N.

maximum force of friction (μ ≈ 1.0) between the person and the ground, if the person has a mass of 70 kg  = 1.0 × 70 × 9.8N = 686 N.

Hence, the force of the storm is (294/686) = 0.42 times typical maximum force of friction.

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Answer:

C. A glass of lemonade starting to freeze in a warm room.

Explanation:

This is because there is no source of cold to properly freeze the lemonade, freezing happens when a cold source starts to slow down the movement of atoms. So if there's nothing there to freeze it, it can't freeze.

Alternatively, there are some materials that harden when not subjected to heat, but that's not freezing. It's just cooling.

The difference in pressure between the wide and narrow ends of the tube is 2102.96 Pa.

The difference in pressure between the wide and narrow ends of the tube if an incompressible liquid is flowing through a tube that suddenly narrows and increases its velocity is calculated as follows. We have to apply Bernoulli's equation to find the difference in pressure.Bernoulli's equation:P1 + 0.5 ρ v1^2 = P2 + 0.5 ρ v2^2P1 and P2 represent the pressure at points 1 and 2, respectively. ρ is the liquid's density, while v1 and v2 are the liquid's velocity at points 1 and 2, respectively.

The pressure difference is:P1 - P2 = (1/2) ρ (v2^2 - v1^2)P1 is the pressure at the wide end of the tube, which is equivalent to the ambient pressure, which we'll take as 1 atm. The velocity at the wide end of the tube, v1, is 1.4 m/s. The velocity at the narrow end of the tube, v2, is 3.2 m/s. Density, ρ, is equal to 1065 kg/m³, as mentioned in the question.

P1 - P2 = (1/2) ρ (v2^2 - v1^2)P1 - P2 = (1/2) (1065 kg/m³) (3.2 m/s)^2 - (1.4 m/s)^2P1 - P2 = 3028.62 Pa - 925.66 PaP1 - P2 = 2102.96 Pa.

Therefore, the difference in pressure between the wide and narrow ends of the tube is 2102.96 Pa.An incompressible liquid is a fluid that does not compress significantly and is therefore not affected by pressure changes.

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The time constant (in ms) of the RC circuit is 3.75 ms. Hence, the correct option is  (d) 3.75 ms.

The rate of decay of the current in a charging capacitor is proportional to the current in the circuit at that time. Therefore, it takes longer for a larger current to decay than for a smaller current to decay in a charging capacitor.A capacitor is discharged through a 20.0 Ω resistor.

The discharge current decreases to 22.0% of its initial value in 1.50 ms. We can obtain the time constant of the RC circuit using the following formula:$$I=I_{o} e^{-t / \tau}$$Where, I = instantaneous current Io = initial current t = time constant R = resistance of the circuit C = capacitance of the circuit

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The time constant of the RC circuit is approximately 0.674 m s.

To determine the time constant (τ) of an RC circuit, we can use the formula:

τ = RC

Given that the discharge current decreases to 22.0% of its initial value in 1.50 m s, we can calculate the time constant as follows:

The percentage of the initial current remaining after time t is given by the equation:

I(t) =\(I_oe^{(-t/\tau)\)

Where:

I(t) = current at time t

I₀ = initial current

e = Euler's number (approximately 2.71828)

t = time

τ = time constant

We are given that the discharge current decreases to 22.0% of its initial value. Therefore, we can set up the following equation:

0.22 =\(e^{(-1.50/\tau)\)

To solve for τ, we can take the natural logarithm (ln) of both sides:

ln(0.22) = \(\frac{-1.50}{\tau}\)

Rearranging the equation to solve for τ:

τ = \(\frac{-1.50 }{ ln(0.22)}\)

Calculating this expression:

τ ≈ 0.674 m s

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The relationship between kinetic energy and speed is directly proportional.

The given problem is based on the Kinetic energy. The energy possessed by any object by virtue of its motion is known as kinetic energy of an object. The mathematical expression for the kinetic energy is given as,

\(KE = \dfrac{1}{2}mv^{2}\)

Here,

m is the mass of object.

v is the speed of object.

the mass of any object remains constant. Then, the kinetic energy is given as,

\(KE \propto v^{2}\)

Thus, we can conclude that the kinetic energy is directly proportional to the square of speed of object.

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We know

\(\boxed{\sf Power=\dfrac{Work\:done}{Time}}\)

\(\\ \sf\longmapsto Power=\dfrac{Force\times displacement}{Time}\)

\(\\ \sf\longmapsto Power=\dfrac{50\times 10}{10}\)

\(\\ \sf\longmapsto Power=\dfrac{500}{10}\)

\(\\ \sf\longmapsto Power=50W\)

ANSWER:

294 N

STEP-BY-STEP EXPLANATION:

We can calculate the weight as follows:

Replacing:

Weight is 294 N

If you find yourself being hit by infrared rays, you are experiencing exposure to infrared radiation.

Infrared radiation refers to the electromagnetic waves with longer wavelengths than those of visible light. These rays are not visible to the human eye but can be detected by specialized devices or sensors. When you feel the sensation of being hit or exposed to infrared rays, it means that your body is absorbing this form of radiation.

Infrared radiation is commonly emitted by warm objects, such as the sun, fire, or heated surfaces. It is important to note that excessive or prolonged exposure to intense infrared radiation can have detrimental effects on human health, including thermal burns and tissue damage. Therefore, it is advisable to take appropriate precautions and limit exposure to high-intensity infrared sources.

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Complete Question:

A long distance runner running a 5.0km track is pacing himself by running 4.5km at 9.0km/hr and the rest at 12.5km/hr. What is the average speed?​

Answer:

Average speed = 9.7333 km/h

Explanation:

Given the following data;

To find the average speed;

First of all, we would determine the time taken to cover distance A in speed A by using the formula;

\( Time \ A = \frac {Distance \; A}{Speed \; A} \)

Substituting the values into the formula, we have;

\( Time \ A = \frac {4.5}{9.5} \)

Time A = 0.4737 hours

Total distance = distance A + distance B

5 = 4.5 + distance B

Distance B = 5 - 4.5

Distance B = 0.5 Km

Next, we would determine the time to cover distance B in speed B;

\( Time \ B = \frac {0.5}{12.5} \)

Time A = 0.04 hours

Total time = time A + time B

Total time = 0.4737 + 0.04

Total time = 0.5137 hours

Now, we would solve for the average speed;

Mathematically, the average speed of an object is given by the formula;

\( Average \; speed = \frac {total \; distance}{total \; time} \)

\( Average \; speed = \frac {5}{0.5137} \)

Average speed = 9.7333 km/h

Three different people weigh a standard mass of 2.00 g on the same balance. Each person obtains a reading of 7.32 g for the mass of the standard. These results imply that the balance that was used is neither accurate nor precise.

The question states that three different people weigh a standard mass of 2.00 g on the same balance. However, each person obtains a reading of 7.32 g for the mass of the standard. This result is unexpected because the actual mass of the standard is known to be 2.00 g.

If the balance were accurate and functioning correctly, each person should have obtained a reading close to the actual mass of 2.00 g. The fact that all three individuals obtained the same reading of 7.32 g suggests that there is a systematic error or issue with the balance.

There are several possible explanations for the discrepancy:

1. Calibration Error: The balance may be improperly calibrated. Calibration involves adjusting the balance to ensure accurate measurements. If the balance is not calibrated correctly, it can lead to inaccurate readings.

2. Zero Error: The balance might have a zero error, meaning that it does not read zero when there is no mass on the pan. This can cause all subsequent measurements to be incorrect by a consistent amount.

3. Faulty Measurement Mechanism: The internal components of the balance responsible for measuring mass, such as the springs or sensors, may be malfunctioning or damaged. This can lead to inaccurate readings.

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Answer:

a and c

Explanation:

The rate of acceleration is 2 m/s² and correct option is d.

To find the rate of acceleration, we can use the formula for acceleration:

Acceleration (a) = (Change in velocity) / (Change in time)

Given:

Initial velocity (u) = 6 m/s

Final velocity (v) = 10 m/s

Time interval (t) = 2 s

Change in velocity = v - u = 10 m/s - 6 m/s = 4 m/s

Plugging these values into the formula for acceleration:

Acceleration (a) = (Change in velocity) / (Change in time)

= 4 m/s / 2 s

= 2 m/s²

Therefore, the rate of acceleration is 2 m/s². So the correct answer is d. 2 m/s².

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The oil slick will be covering an area of 16500 km² approximately. Given that;1 barrel ≈ 0.15899 m³and 1 Tonne ≈ 1 m³ (as 1 Ton of crude oil weighs 1000 kg). Therefore, 1 barrel ≈ 0.15899 Tonne ≈ 0.15899 m³Assuming the volume of oil spreads out smoothly to a slick of one molecule thick.

Then the volume of the slick would be equal to the total volume of the oil spilled. Volume of oil spilled = 4100 × 1000 Tonne Volume of oil spilled = 4100000 m³Area covered by the slick can be obtained as; Area of slick = volume of slick / thickness of slick.

Thickness of slick can be taken as 1 molecule = 1 × 10⁻⁹ m Area of slick = (4100 × 1000) / (1 × 10⁻⁹)Area of slick ≈ 4.1 × 10¹² m²Area of slick = 4.1 × 10⁶ km² (as 1 km² = 10⁶ m²)Area of slick ≈ 16500 km² (to two significant figures)Therefore, the oil slick will be covering an area of 16500 km² approximately.

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Answer:

E=mc△T

E=0.012(4200)(99-98)

E=50.4J

50.4 joules of heat is used

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Question 4 : 770 mm Hg = how many pascals

Conversion from mm Hg to atm :

Conversion from atm to kPa :

Conversion of kPa to Pa :

hence, 770 mm Hg = 1.026 × 10⁵ Pa

Two objects having the same momentum do not necessarily have the same velocities in the same direction.

This is because momentum is a vector quantity, which means that it has both magnitude and direction. Velocity, on the other hand, is also a vector quantity since it has magnitude and direction. Therefore, two objects can have the same momentum, but their velocities can be different in terms of direction or magnitude or both. The momentum of an object is the product of its mass and its velocity. When two objects of different masses and different velocities are compared, the one with the larger mass and slower velocity has more momentum than the one with smaller mass and faster velocity.

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Question :-

Answer :-

Explanation :-

As per the provided information in the given question, we have been given that the Speed of scooter is 15 m/s . And, we have been asked to calculate the Distance , covered by the scooter in 45 seconds .

For calculating the Distance , we will use the Formula :-

\( \bigstar \: \: \boxed{ \sf{ \: Distance \: = \: Speed \: \times \: Time \: }} \)

Therefore , by Substituting the given values in the above Formula :-

\( \dag \: \: \: \sf { Distance \: = \: Speed \: \times \: Time \: } \)

\( \longmapsto \: \: \: \sf { Distance \: = \: 15 \: \times \: 45 \: } \)

\( \longmapsto \: \: \textbf {\textsf { Distance \: = \: 675 meter}} \)

\( \underline {\rule {180pt} {4pt}} \)

In problem 3, the artillery shell takes approximately 35.08 seconds to reach its maximum range. At 2/3 of the total time, the x-coordinate is approximately 6511.77 meters and the y-coordinate is approximately 1136.17 meters.

In problem 4, the ball will fall approximately 1.91 meters vertically by the time it reaches home plate, which is 19 meters away.

Problem 3:

To calculate the time it takes for the artillery shell to reach its maximum range, we use the equation:

\(t_m_a_x\) = (2 * v * sin(θ)) / g

Substituting the given values:

\(t_m_a_x\) = (2 * 300.0 * sin(55.0°)) / 9.8

\(t_m_a_x\) ≈ 35.08 seconds

The total time of motion is twice the time to reach the maximum height:

\(t_t_o_t_a_l\)= 2 *\(t_m_a_x\)

\(t_t_o_t_a_l\) ≈ 70.16 seconds

To find the x and y coordinates at 2/3 of the total time, we use the equations:

x = v * cos(θ) * t

y = v * sin(θ) * t - (1/2) * g * t²

Substituting t = (2/3) * \(t_t_o_t_a_l\):

x = 300.0 * cos(55.0°) * (2/3) * 70.16

x ≈ 6511.77 meters

y = 300.0 * sin(55.0°) * (2/3) * 70.16 - (1/2) * 9.8 * ((2/3) * 70.16)²

y ≈ 1136.17 meters

Therefore, at 2/3 of the total time, the x-coordinate is approximately 6511.77 meters, and the y-coordinate is approximately 1136.17 meters.

Problem 4:

Since the pitch is thrown horizontally, the vertical displacement is determined solely by the effect of gravity. The equation for vertical displacement is:

y = (1/2) * g * t²

To find the time taken to travel the horizontal distance of 19 meters:

t = d / v

t = 19 / 30.0

t ≈ 0.633 seconds

Substituting the time into the equation for vertical displacement:

y = (1/2) * 9.8 * (0.633)²

y ≈ 1.91 meters

Therefore, by the time the ball reaches home plate, it will fall approximately 1.91 meters vertically.

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Answer:

(3) -

(4) - \(\Delta \vec y =1.963 \ m\)

Explanation:

Answer the given projectile motion problems.

Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone, without any other forces acting on it. It follows a curved path known as a parabola. Here are the key points about projectile motion:

Understanding these principles helps in predicting and analyzing the motion of objects such as projectiles, including projectiles launched in sports, cannonballs, or even objects in space.

We will be utilizing the following four kinematic equations:

\(\boxed{\left\begin{array}{ccc}\text{\underline{The 4 Kinematic Equations:}}\\\\1. \ \vec v_f=\vec v_0+\vec at\\\\2. \ \Delta \vec x=\frac{1}{2}(\vec v_f-\vec v_0)t\\\\3. \ \Delta \vec x=\vec v_0t+\frac{1}{2}\vec at^2\\\\ 4. \ \vec v_f^2=\vec v_0^2+2\vec a \Delta \vec x \end{array}\right}\)

We will also use the range formula, which is a formula we can use in a special case:

\(\boxed{\left\begin{array}{ccc}\text{\underline{The Range Formula:}}\\\\R=\dfrac{v_0^2\sin(2\theta)}{g} \end{array}\right}\)

The range formula allows us to calculate the horizontal distance an object travels. But we may only use this if the object lands at the same height it is initially fired at.

I personally like making a table. Where I can split up horizontal and vertical components up and write the information I know.

Here is a template:

\(\boxed{\left\begin{array}{ccc}\text{\underline{Horizontal Components:}}\\\\\vec v_{x}=\vec v\cos(\theta)\\ \\ \vec a_x=0 \ m/s^2 \\ \\ \Delta \vec x= \ ?? \ m \\ \\ t= \ ?? \ s\end{array}\right} \ \ \boxed{\left\begin{array}{ccc}\text{\underline{Vertical Components:}}\\\\\vec v_{y}=\vec v\sin(\theta)\\ \\ \vec a_y=-9.8 \ m/s^2 \\ \\ \Delta \vec y= \ ?? \ m \\\\ t=\ ?? \ s \end{array}\right}\)\(\hrulefill\)

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