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Helga [31]
3 years ago
3

Consider two cars moving along the same straight road in opposite directions. Car A has a mass of 500kg and has a constant speed

of 20m/s; car B has a mass of 800kg and a constant speed of 15m/s. What can you say about the net forces on the cars? 1. Car A experiences greater net force than car B.2. Car B experiences greater net force than car A.3. Both cars experience equal net forces.
Physics
2 answers:
Y_Kistochka [10]3 years ago
8 0

Answer:

3. Both cars experience equal net forces.

Explanation:

According to Newton's 1st law, object subjected to 0 force, or net force equals 0, would be at rest or maintain a constant speed. In our cases both cars are at constant speed (whatever they are) so each of their individual net force must be 0. Both cars experience equal net forces.

Finger [1]3 years ago
3 0

Answer:

3. Both cars experience equal net forces

Explanation:

The net force (∑F) on a body, according to Newton's second law of motion, is the product of the mass (m) of the body and the acceleration (a) of motion of the body. This can be written mathematically as follows;

∑F = m x a       -----------------------(i)

Given;

Car A => mass = 500kg and constant speed = 20m/s

Car B => mass = 800kg and constant speed = 15m/s

Both cars are moving at constant speed. This means that the velocities of the two cars are not changing. i.e

For Car A, initial velocity (u) = 20m/s and final velocity (v) =  20m/s. Hence acceleration which is a time rate of change in velocity is zero;

i.e

a = (v - u) / t

a = (20 - 20) / t = 0

For Car B, initial velocity (u) = 15m/s and final velocity (v) =  15m/s. Hence acceleration which is a time rate of change in velocity is zero;

i.e

a = (v - u) / t

a = (15 - 15) / t = 0

Therefore, the accelerations of both cars are zero (0). In other words, constant speed means zero acceleration.

Now, let's calculate the net force on Car A;

∑F = m x a        [where m = 500kg, a = 0]

∑F = 500 x 0

∑F = 0 N

Also, let's calculate the net force on Car B;

∑F = m x a        [where m = 800kg, a = 0]

∑F = 800 x 0

∑F = 0 N

Therefore, both cars experience equal net forces of zero (0);

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4 years ago
a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik
grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:  

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)  

Where:  

y=0 is the height of the ball when it hits the ground  

y_{o}=70 m is the initial height of the ball

V_{o}=82m/s is the initial velocity of the ball  

t is the time when the ball strikes the ground

g=9.8m/s^{2} is the acceleration due to gravity  

Having this clear, let's find t from (1):  

0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2} (2)  

Rewritting (2):

-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0 (3)  

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}  (4)

Where:

a=-\frac{1}{2}(9.8m/s^{2}

b=82m/s

c=70m

Substituting the known values:

t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}  (5)

Solving (5) we find the positive result is:

t=17.68 s

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schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

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Explanation:

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