When you cut a object does the density same the same between the two objects or no and if no what will happen

A car is moving with a constant velocity of 25 m/s. Which of the following is true?

S_A_V [24]

b) the net force on the car is zero.

Explanation:

Let's analyze each option one by one:

a) the force from the engine is greater than all the forces of friction. --> FALSE. In fact, the car is moving at constant velocity: this means that its acceleration is zero,

a = 0

and so Newton's second law becomes

$\sum F = ma = 0$

where $\sum F$ is the net force on the car and m is its mass. This means that the net force on the car is zero: so, the force from the engine cannot be greater than all the forces of friction, otherwise the net force cannot be zero.

b) the net force on the car is zero. --> TRUE, for what we said at point A)

c) the inertia is changing. --> FALSE. The inertia of an object just depend on the mass and the velocity of the object: as neither the mass nor the velocity are changing in this problem, then the inertia of the car is not changing.

d) the forces of friction are proportional to the acceleration of the car. --> FALSE. Generally, the force of friction acting on an object moving on a flat surface is

$F_f = \mu mg$

where $\mu$ is the coefficient of friction, m is the mass, and g the acceleration of gravity. Therefore, the force of friction does not depend on the acceleration of the car.

e) All of the above. --> FALSE

Learn more about net force and Newton's second law:

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6 0

3 years ago

John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

8 0

3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

3 years ago

Read 2 more answers

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

2 years ago

Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can

svetlana [45]

Answer:

3.6 m

Explanation:

$\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \ 0.25 \ mm\\\\= 0.25 *10^{-3} m$

$\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\$

Also

$\beta = 1 \ mm \ fringe \ width$

$D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} = 3.6 m$

Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m

4 0

3 years ago