They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.
Hope this helps :)
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
Answer:
A). A virtual image cannot be formed on a screen.
Explanation:
A virtual image can not be formed on a screen.
For image:
1.A virtual image can be viewed by the unaided eye.
2. A real image must be erect or maybe inverted.
3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.
4.A virtual image can be photographed.
So the option A is correct.
Okay, haven't done physics in years, let's see if I remember this.
So Coulomb's Law states that

so if we double the charge on

and double the distance to

we plug these into the equation to find
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So we see the new force is exactly 1/2 of the old force so your answer should be

if I can remember my physics correctly.
Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped:
![M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]](https://tex.z-dn.net/?f=M-m%20%3D%20%5Cfrac%7BF%7D%7Bg-a%7D%20-%20%5Cfrac%7BF%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B1%7D%7Bg-a%7D%20-%20%5Cfrac%7B1%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%28g%2B%28a%2F2%29%29%20-%20%28g-a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7Bg%2B%28a%2F2%29%20-%20g%20%2B%20a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%283a%2F2%29%7D%7Bg%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7Bag%7D%7B2%7D%7D%20%5D)