A simple pendulum of length L has a period of T seconds. What should be the length for the period to be 3Tseconds?

Answer:

Explanation:

Very

Answer:Because of the relative sizes of the moon and sun and their relative distances from Earth,

Explanation:

Answer:

x=6.39 m

Explanation:

Rectangular Components of a Vector

Vectors can be expressed in several forms, including the magnitude-angle form, and the rectangular-coordinates form.

If the magnitude r and angle β are given, we can find the rectangular coordinates (x,y) as follows:

\(x=r\cos\beta\)

\(y=r\sin\beta\)

The vector of the question has a magnitude of r=9.55 m and an angle β=-48°. The x-component is:

\(x=9.55\cdot\cos(-48^\circ)\)

Using a digital calculator:

x=6.39 m

Answer:

Twice as much

Explanation:

D 5-7 Be the only person I know of is the Book with all men and men and people experience the first of the wisdom and

The acceleration produced will be one-half as much.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. SI unit of acceleration is meter/second² (m/s²).

From Newton's 2nd law of motion; it can be written that:

Force = mass × acceleration

acceleration = force/mass.

Now, according to the question, an unbalanced force of 2 newtons applied to a given mass produces an acceleration.

So, when  an unbalanced force of 1 newton is applied to the same mass, the force applied on the object becomes half.

Hence, from the equation, the acceleration produced will be one-half as much.

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Answer: Distance traveled in time t is s

Explanation: Self Explanatory

According to the question the acceleration of the crate from Newton's Second Law is 0.5m/s2.

Acceleration is the rate of change of velocity of an object over a period of time. It is a vector quantity, meaning it has both a magnitude and a direction. Acceleration is typically denoted by the letter "a". Acceleration can be caused by a variety of forces, such as gravity, friction, or a net external force. When an object is accelerating, its velocity is changing in magnitude or direction, or both.

A- The acceleration of the crate from Newton's Second Law is given by the equation F=ma, where F is the force applied, m is the mass of the crate and a is the acceleration. Substituting F=50N and m=100kg, we get a = F/m = 50N/100kg = 0.5m/s2.

B- The distance travelled in 4s can be calculated using the equation s = ut + 0.5at2. Since the crate starts from rest, its initial velocity (u) is 0. Substituting u = 0, a = 0.5m/s2 and t = 4s, we get s = 0 + 0.5 x 0.5 x 42 = 8m.

C- The work done can be calculated using the equation W = Fs, where F is the force applied and s is the distance travelled. Substituting F = 50N and s = 8m, we get W = 50N x 8m = 400J.

D- The velocity of the crate at the end of 4s can be calculated using the equation v = u + at, where u is the initial velocity, a is the acceleration and t is the time. Substituting u = 0, a = 0.5m/s2 and t = 4s, we get v = 0 + 0.5 x 4 = 2m/s.

E- The kinetic energy of the crate at the end of 4s can be calculated using the equation KE = 0.5mv2, where m is the mass of the crate and v is the velocity. Substituting m = 100kg and v = 2m/s, we get KE = 0.5 x 100kg x 22 = 200J.

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Answer:

Answer:

F = M a       Newton's second law

Γ = I α         Corresponding law for circular motion

α = 20 * 2 π rad / sec / 10 sec = 4 π rad / sec^2

α = 12.6 /sec^2       (rad not an actual unit)

Γ = 5E-3 * 12.6E1 = .63 kg-m^2 / sec^2

(5E-3 / 12.6 = 4E-4 = 4xxE-2   seems like a mistake made in arithmetic)

Note:   kg-m^2 * 1 / sec^2 = kg-m^2 / sec^2 = N-m

One needs to "multiply" inertia * rad/sec^2 to get required units of torque

A minispherical cow is thrown upwards with an initial velocity of 7.0 m/s; the earth's acceleration due to gravity is -9.8 m/s, and the cow's height after 0.73 s is approximately 2.64 m above its initial position.

h = \(V_i\)t + (1/2)g\(t^2\)

(Here, h = the height of the cow above its initial position \(V_i\)= the initial velocity of the cow that is 7.0 m/s upwards, g = acceleration due to gravity that is -9.8 m/\(s^2\) downwards, t = the time since the cow was thrown)

putting in the given values, one can get:

h = (7.0 m/s)(0.73 s) + (1/2)(-9.8 m/\(s^2\))\((0.73 s)^2\)

h ≈ 2.64 meters

Hence, the cow's height after 0.73 seconds is approximately 2.64 meters above its initial position .

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Answer: 12 ounce steak on venus

Explanation: venus = closer to the sun. The sun = more potential gravity, which would make it heavier than a quarter pound hamburger on jupiter.

The caravan has an initial speed of

85 km/h × (1000 m/km) × (1/3600 h/s) ≈ 23.611 m/s

If it slows down at a constant rate, then it will cover a distance x such that

0² - (23.611 m/s)² = 2 (-1.3 m/s²) x

Solve for x :

x = -(23.611 m/s)² / (2 (-1.3 m/s²)) ≈ 214.417 m ≈ 210 m

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

work done in pumping the entire fuel is 466587 J

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical, we assume that the fuel within also takes the cylindrical shape.

Also, we assume that the fuel completely fills the tank.

volume of a cylinder = \(\pi r^{2}l\)

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x \(1.5^{2}\) x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = 466587 J

Answer:

   r = <2.640 10⁶, 1.01 10⁸, 0> m

Explanation:

For this exercise we are going to solve it for each direction separately,

we locate a fixed reference frame in space at the height of the rocket, such that the position of the rocket is

  r₀ = <4, 4, 0> 10³ m

X axis

the initial velocity of the ship on this axis is v₀ₓ = 0, when it passes through the point x₀ = 4 km it ignites the rockets, experiencing a force of Fₓ = 7.0 10⁵ N for 21 s and the rockets turn off

They ask where it is after one hour t = 1 h = 3600 s

Let's apply Newton's second law

        Fₓ = m aₓ

        aₓ = Fₓ / m

        aₓ = 7.0 10⁵/2 10⁴

        aₓ = 3.5 10¹ m / s

Let's use kinematics to find the distance

for the first t₁ = 21 s the movement is accelerated

        x₁ = x₀ + v₀ t₁ + ½ aₓ t₁²

        x₁ = x₀ + ½ aₓ t₁²

        x₁ = 4000 + ½ 35 21² = 4000 + 7717,.5

        x₁ = 11717.5 m

this instant has a speed of

        vₓ = v₀ₓ + aₓ t

        vₓ = aₓ t ₁

        vₓ = 35  21

        vₓ = 735 m / s

the rest of the time there is no acceleration so it is uniform motion at this speed

        t₂ = 3600 - 21

        t₂ = 3576 s

         vₓ = x₂ / t₂

         x₂ = vₓ t₂

         x₂ = 735  3576

         x₂ = 2,628 10⁶ m

the total distance traveled in this direction is

         x_total = x₁ + x₂

         x_total = 11717.5 + 2.628 10⁶

         x_total = 2,640 10⁶ m

Y axis

on this axis it is in the initial position of y₀ = 4000 m, with an initial velocity of \(v_{oy}\) = 28 10³ m / s and there is no force on this axis F_{y} = 0

The movement in this axis is uniform,

       v_{y} = y / t

       y = v_{y} t

       y = 28 10³/3600

       y = 1.01 10⁸ m

the total distance is

       y_total = y₀ + y

       y_total = 4000 + 1.01 10⁸

       y_total = 1.01 10⁸ m

Z axis

the initial position is z₀ = 0, with an initial velocity of v₀ = 0 and in this axis there is no force F_{z} = 0

the movement is uniform

         z = 0

the final position of the rocket after 1 h is

      r = <2.640 10⁶, 1.01 10⁸, 0> m

Answer:

make a drinkable water filter

find like a bottle or something

put rocks and sands sticks etc

scoop some of the salt water

it might not taste the best but you won't least die or get hydrated

Explanation:

Answer:

24.2 m/s

Explanation:

Mass is irrelevant in this situation....

Displacement:  ( to find time)

 x = xo + vo t - 1/2 at^2

 30= 0   + 0  - 1/2 (9.8)t^2

              t = 2.47 seconds

Velocity:

vf = a t   = 9.8 (2.473)  = 24.2 m/s

Answer: An jack makes changing a tire easier because it lifts up the car to get the tire off of the ground.

Explanation:

Answer:

faster than

Explanation:

because the sun gave it nutrients

The time taken is obtained as 6.95 s.

The time taken could be obtained from the equations of kinematics. In this case, we have to first convert the given velocities to m/s as follows;

u; 70 * 1000/3600 = 19.44 m/s

v; 110  * 1000/3600 = 30.56 m/s

Then we apply the equations of motion;

v = u + at

a =  1.6 m/s2

We now have;

v - u/a = t

t = 30.56 m/s - 19.44 m/s/ 1.6 m/s2

t = 6.95 s

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The number of people you will stack to reach the top of the CN Tower (553 m) is 316 people

We'll begin by converting 175 cm to m. This can be obtained as illustrated below:

100 cm = 1 m

Therefore,

175 cm = (175 cm × 1 m) / 100 cm

175 cm = 1.75 m

Thus, 175 cm is equivalent to 1.75 m

The number of people needed to be stacked to get to the top of the CN tower can be o btained asfollow:

Number of people needed = Height of tower / height of a person

Number of people needed = 553 / 1.75

Number of people needed = 316 people

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The true statement about  gravitational mass is the more mass an object has, the more gravitational acceleration it will experience.

option B.

The gravitational acceleration of an object is the acceleration an object with mass experiences due to impact of force of gravity on the object.

g = ( GM ) / ( R² )

where;

From the equation given above, the acceleration of an object increases with increase in the mass of the object and decrease in the radius of the object.

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answer yuo do not have it

Question:

An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 61.3g.

If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density = 13.6 g/cm3).

Answer:

\(Volume= 4.51cm^3\)

Explanation:

Given

Liquid: Mercury

\(Density = 13.6\frac{g}{cm^3}\) -- of liquid

\(Mass = 61.3g\)

Required

Determine the volume at which the liquid will sink

First, we need to calculate the volume of the sphere for which it will have the same density as the mercury.

This is calculated as follows

\(Density = \frac{Mass}{Volume}\)

Make Volume the subject

\(Volume= \frac{Mass}{Density }\)

Substitute values for Mass and Density

\(Volume= \frac{61.3}{13.6}\)

\(Volume= 4.51cm^3\)

This implies that:

If the volume of the sphere is less than 4.51, the density will be greater than the density of the mercury and so the object will sink

Take for instance:

\(Volume = 4cm^3\)

\(Density = 61.3/4 = 15.325g/cm^3\)

So, fill in the gap with \(4.51cm^3\)

Answer:

C 1 degree

Explanation:

Answer:

Total length of pool \(4.4\) meter

Explanation:

The image for the question is attached for better understanding

\(Tan \Theta= \frac{1.5}{1.2}\\\Theta_a = tan^{-1} (1.25)\\\Theta_a = 51.34\)degree

Using the Snell’s law –

\(n_a * sin\Theta_a = n_b * sin\Theta_b\\1 * sin 51.34 = 1.33 * sin\Theta_b\\sin\Theta_b = 0587\\\Theta_b = 35.94\\Tan \Theta_b = \frac{BC}{4} \\BC = 4 * tan (35.94) \\BC = 2.9\)

Total length of pool \(= 1.5 +2.9 = 4.4\) meter

Answer:

Explanation:

i is angle of incidence and r is angle of refraction .

Tan i = 1.5 / 1.2 = 1.25

i = 51.34⁰

refractive index of water = 1.33

sin i / sinr = 1.33

sin 51.34 / sin r = 1.33

sinr = .78 / 1.33 = .5864

r = 36⁰

From the figure given ,

AB / AP = Tan 36

AB = AP tan 36 = 4 x tan 36 = 2.9 m

D = OA + AB

= 1.5 + 2.9

= 4.4 M

Distance of key from edge = 4.4 m .

Answer:

See below

Explanation:

rho = R A/l     R = resistance   A = cross sectional area l = length

wind (b), also some others are:

- steam

- gas

- liquid fules

The electric field at point B, located at the midpoint between the upper left and right corners of the square, can be approximated as 3.244 x \(10^6\) N/C in the upward direction.

To find the electric field at point B, we can consider the contributions from the two charges placed at the lower corners of the square. Since the charges are the same and the distance to point B is the same for both charges, the magnitudes of the electric fields produced by each charge will be equal.

First, let's calculate the magnitude of the electric field produced by one of the charges at point B using Coulomb's Law:

Electric field due to a point charge (E) = (k * q) / \(r^2\)

Where:

- k is the electrostatic constant, approximately equal to 8.99 x 10^9 \(10^9\)N \(m^2/C^2\)

- q is the charge of the object, 7.25 μC (7.25 x \(10^-^6\) C)

- r is the distance from the charge to point B, which is half the length of the square's side, 0.201 m / 2 = 0.1005 m

Plugging in the values, we have:

E = (8.99 x \(10^9 N m^2/C^2\) * 7.25 x \(10^-^6\) C) / (0.1005 \(m)^2\)

E ≈ 1.622 x \(10^6\) N/C

Since the electric fields from the two charges at the lower corners have equal magnitudes and point in the same direction (upward), the total electric field at point B is twice the magnitude of the individual electric field:

Electric field at point B = 2 * E ≈ 2 * 1.622 x \(10^6\) N/C

Electric field at point B ≈ 3.244 x \(10^6\) N/C

Therefore, the electric field at point B, midway between the upper left and right corners of the square, is approximately 3.244 x \(10^6\)N/C upward.

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