Answer:
45.3°C
Explanation:
Heat gained = mass × specific heat × increase in temperature
q = mC (T − T₀)
Given C = 0.128 J/g/°C, m = 94.0 g, q = 305 J, and T₀ = 20.0°C:
305 J = (94.0 g) (0.128 J/g/°C) (T − 20.0°C)
T = 45.3°C
Grass dear wolf is the right awnser
Answer:
<em>The velocity of the carts after the event is 1 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of the individual momentums:

If a collision occurs and the velocities change to v', the final momentum is:

Since the total momentum is conserved, then:
P = P'
In a system of two masses, the equation simplifies to:

If both masses stick together after the collision at a common speed v', then:

The common velocity after this situation is:

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

The velocity of the carts after the event is 1 m/s
A- by kicking the ball into the net/goal
Answer:
The coefficient of friction and acceleration are 0.37 and 2.2 m/s²
Explanation:
Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.
Given that,
Mass of block = 60 kg
Acceleration = 2.0 m/s²
Mass = 100 kg
Horizontal force = 340 N
Let the frictional force be f.
We need to calculate the frictional force
Using balance equation

Put the value into the formula



We need to calculate the coefficient of friction
Using formula of friction force




We need to calculate the acceleration of the 100 kg block
Using formula of newton's law




Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²