Ask question

Ask question

All categories

2 years ago

5

A circus truck full of monkeys has a mass of 3175 kg. The kinetic energy of the circus truck is 128,587.5 J. Calculate the speed

of the truck.

1 answer:

valentina_108 [34]2 years ago

7 0

Explanation:

please mark me as brainlest

You might be interested in

Which has the most momentum?

boyakko [2]

Answer:

Both objects have the same magnitude of momentum.

Explanation:

If an object of mass $m$ is moving at a velocity of $v$ , the momentum of that object would be $m\, v$ .

The $100\; {\rm g}$ ( $0.1\; {\rm kg}$ ) object is moving at a speed of $1\; {\rm m\cdot s^{-1}}$ . The magnitude of the momentum of this object would be $0.1\; {\rm kg} \times 1\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Similarly, the momentum of the $1\; {\rm g}$ ( $10^{-3}\; {\rm kg}$ ) object moving at a speed of $100\; {\rm m\cdot s^{-1}}$ would be $10^{-3}\; {\rm kg} \times 100\; {\rm m\cdot s^{-1}} = 0.1\; {\rm kg \cdot m \cdot s^{-1}}$ .

Hence, the magnitude of momentum is the same for the two objects.

7 0

1 year ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

Which of the following would be most likely to have individual characteristics?

ZanzabumX [31]

Answer:

D. Fingerprints

Explanation:

answers for Plato. I took the test

3 0

3 years ago

Higher temperatures along with moisture climates generally increase the rate of

Bad White [126]

Answer:

Option (D)

Explanation:

Weathering refers to the disintegration of rocks due to the different geological processes that are initiated by agents such as wind, water and ice.

In a region where the temperature is high, and there is sufficient amount of moisture content in the air, there occurs high rate of physical, chemical and biological type of weathering.

Physical weathering processes such as freeze-thaw method and exfoliation method commonly occurs in this type of region.
Chemical weathering processes such as hydrolysis, oxidation, as well as carbonation method occurs in the region of high temperature, moist climatic condition.
Biological weathering also is possible up to some extent as the plants takes up water from the rainfall and also receive enough sunlight for its growth, as a result of which it leads to the breakdown of rocks.

Thus, the correct answer is option (D).

4 0

3 years ago

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

marta [7]

1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

$u = 27.7 m/s$ --> initial velocity of the car

$\mu=0.804$ --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

$F=-\mu mg$

where m is the car's mass and $g=9.81 m/s^2$ is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

$a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2$

And the negative sign means it is a deceleration.

2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

$v^2 - u^2 = 2aS$

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

$S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m$

3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

$d_t = ut=(27.7 m/s)(0.543 s)=15.0 m$

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

$d_b=48.6 m$

So the total distance your car covers is

$S'=d_t+d_b=15.0 m +48.6 m=63.6 m$

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

$d=S'-S=63.6 m-48.6 m=15.0 m$

6 0

3 years ago