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3 years ago

13

Which car has the greater momentum?

1 answer:

Marianna [84]3 years ago

6 0

The answer is C. p=mv p=18x30 = 540kgm/s

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Kieran takes off from rest down a 50 m high, 10° slope on his jet-powered skis. The skis have a thrust of 280 N parallel to the

xz_007 [3.2K]

Answer:0.46

Explanation:

Given

Initial height $h=50 m$

inclination $\theta =10^{\circ}$

Thrust $=280 N$

combined mass of kieran and skis $m=50 kg$

Speed at the bottom $v=40 m/s$

From Work Energy Theorem

Work done by all the force is equal to change in kinetic Energy

$W_{gravity}+W_{friction}+W_{thrust}=\frac{1}{2}mv^2-0$ ------------1

distance traveled along the slope $x=\frac{50}{\sin 10}=287.93 m$

$W_{gravity}=mgh=50\times 9.8\times 50=24500 J$

$W_{thrust}=F\times x=280\times 287.93=80,620.4 J$

$W_{friction}=-\mu mg\cos 10$

substitute in 1

$24,500+80,620.4+W_{friction}=\frac{1}{2}\times 50\times 1600$

$W_{friction}=40,000-24,500-80,620.4$

$-\mu \cdot 50\times 9.8\times 287.93=-65,120.4$

$\mu =\frac{65,120.4}{141,085.7}=0.46$

8 0

2 years ago

Which of the following properties of materials does NOT affect resistance?

Shalnov [3]

Answer:

B thickness

Explanation:

thickness

4 0

2 years ago

What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?

Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0

2 years ago

A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time int

vagabundo [1.1K]

Answer:

16

Explanation:

$\frac{1}{2} \times 2 \times {4}^{2} = 16$

7 0

3 years ago

A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?

dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

$I = MR^2$

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

$\Omega = \frac{Mgd}{I\omega}$

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

$\omega$ = Angular velocity

Replacing the value for moment of inertia

$\Omega= \frac{MgR}{MR^2 \omega}$

$\Omega = \frac{g}{R\omega}$

The value for our angular velocity is not in SI, then

$\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})$

$\omega = 104.7rad/s$

Replacing our values we have that

$\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}$

$\Omega = 1.17rad/s$

The precession frequency is

$\Omega = \frac{2\pi rad}{T}$

$T = \frac{2\pi rad}{\Omega}$

$T = \frac{2\pi}{1.17}$

$T = 5.4 s$

Therefore the precession period is 5.4s

7 0

2 years ago