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BlackZzzverrR [31]
4 years ago
12

Two trucks with equal mass are attracted to each other with a gravitational force of 5.3 x 10 -4 N. The trucks are separated by

a distance of 2.6 m. What is the mass of either of the trucks?
Physics
1 answer:
HACTEHA [7]4 years ago
3 0

Answer:

7327 kg or 7.3 tons

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408*10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M = M_1 = M_2 is the masses of the 2 objects. and R = 2.6m is the distance between them.

F_G = 5.3*10^{-4}N is the attraction force.

5.3*10^{-4} = 6.67408*10^{-11}\frac{M^2}{2.6^2}

7941265 = \frac{M^2}{2.6^2}

M^2 = 53682948.76

M = \sqrt{53682948.76} \approx 7327 kg or 7.3 tons

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The driver of a car traveling at 22.8 m/s applies the brakes and undergoes a constant deceleration of 2.95 m/s 2 . How many revo
a_sh-v [17]

Answer:

70 revolutions

Explanation:

We can start by the time it takes for the driver to come from 22.8m/s to full rest:

t = \Delta v/a = (22.8 - 0)/2.95 = 7.73 s

The tire angular velocity before stopping is:

\omega_0 = v/r = 22.8 / 0.2 = 114 rad/s

Also its angular decceleration:

\alpha = a / r = 2.95/0.2 = 14.75 rad/s^2

Using the following equation motion we can findout the angle it makes during the deceleration:

\omega^2 - \omega_0^2 = 2\alpha\Delta \theta

where \omega = 0 m/s is the final angular velocity of the car when it stops, \omega_0 = 114rad/s is the initial angular velocity of the car \alpha = 14.75 rad/s2 is the deceleration of the can, and \Delta \theta is the angular distance traveled, which we care looking for:

-114^2 = 2*(-14.75)*\Delta \theta

\Delta \theta = 440rad or 440/2π = 70 revelutions

4 0
3 years ago
2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotatin
ycow [4]

Answer:

\frac{0.065}{r}

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

v=\sqrt{\mu gr}....................(1)

where \mu is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = 9.81m/s^2

r = ?

\mu=?

In order to solve for \mu, we can simply make it the subject of formula from equation (1) as follows;

v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

\mu=\frac{0.8^2}{9.81r}

\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}

8 0
3 years ago
The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector
xz_007 [3.2K]

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components  are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

3 0
3 years ago
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