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Katarina [22]
3 years ago
10

Two cars having different weights are traveling on a level surface at different constant velocities. Within the same time interv

al, greater force will always be required to stop a car that has the greater
a)weight
b)velocity
c)kinetic energy
d)momentum
Physics
1 answer:
GuDViN [60]3 years ago
4 0

I think you're saying that once you start pushing on the cars, you want to be able to stop each one in the same time. 

This is sneaky.  At first, I thought it must be both 'c' and 'd'.  But it's not
kinetic energy, for reasons I'm not ambitious enough to go into.
(And besides, there's no great honor awarded around here for explaining
why any given choice is NOT the answer.)

The answer is momentum.

Momentum is (mass x speed).  Change in momentum is (force x time).

No matter the weight (mass) or speed of the car, the one with the greater
momentum is always the one that will require the greater (force x time)
to stop it.  If the time is the same for any car, then more momentum
will always require more force.


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A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N

We can find the static fricton force as follow,

F=\mu_s * N

Where N is the normal force (mg)

F=0.3*50*9.8\\F=147N

Static friction force at 147N is greater than the force applied hence body does not move.

F=\mu_k N = 0.2*50*9.8= 98N

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3 years ago
A Transformer has a primary coil with 650 turns if the output voltage is 500 volts and an input voltage of 200 volts how many te
maxonik [38]

Answer:

1625

Explanation:

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Ns / 650 = 500/200

Ns = 2.5 x 650

Ns = 1625

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3 years ago
Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,
makkiz [27]

Answer: \theta=cos^{-1}0.991=7.69^o

The following vectors have been given: \vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}

The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54

||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}

cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

7 0
3 years ago
The weight of the atmosphere above 1 m- of
mars1129 [50]

Answer:

1.09 kg.m

Explanation:

no need

4 0
2 years ago
6
Arturiano [62]

Explanation:

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