Answer:
<u>FALSE.</u>
Explanation:
Newton's third law states that :
- <em>Every action has equal and opposite reaction</em>
- <em>That is , the magnitude is the same but the directions are opposite</em>
- <em>The action reaction forces DONOT operate on the same body.</em>
For example ,
If a block is kept on the ground , the action force is the normal force acting on it due to the ground. <em>BUT , NOTE THAT : the reaction force isn't the gravitational force on the body ! It is the normal force acting on the ground due to the block !</em>
Thus,
we conclude that action and reaction forces donot act on the same body and therefore , this case has the <u>answer : FALSE </u>
Answer: 750Kg
Explanation:
Recall that force is the product of the mass M, of an object moving at a uniform acceleration.
i.e Force = Mass x Acceleration
In this case, Mass = ?
Force = 3.00 x 10^3 N = (3.00 x 1000N)
= 3000N
Uniform acceleration = 4.00m/s^2
Force = Mass x Acceleration
3000N = Mass x 4.00m/s^2
Mass = (3000N/4.00m/s^2)
Mass = 750Kg (The SI unit of mass is kilograms)
Thus, the mass of the car is 750Kg
Answer:
B. has a smaller frequency
C. travels at the same speed
Explanation:
The wording of the question is a bit confusing, it should be short/long for wavelength and low/high for frequency. I assume low wavelength mean short wavelength.
All sound wave travel with the same velocity(343m/s) so wavelength doesn't influence its speed at all. It won't be faster or slower, it will have the same speed.
Velocity is a product of wavelength and frequency. So, a long-wavelength sound wave should have a lower frequency.
The option should be:
A. travels slower -->false
B. has a smaller frequency -->true
C. travels at the same speed --->true
D. has a higher frequency --->false
E. travels faster has the same frequency --->false
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
