Answer:
Having a bigger angle above the horizontal
Explanation:
Applying the energy conservation theorem:

The kinetic energy is reduced because of the work done by the friction force.
The friction force is given by:

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

So when the angle of the slide is bigger, the friction force decreases, for example:
for 45 degrees:

for 75 degrees:

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.
If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.
When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.
a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s
Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ
Answer:
Force required to accelerate = 794.44 N
Explanation:
Force required = Mass of horse x Acceleration of horse
Mass of horse and rider, m= 572 kg
Acceleration of horse and rider, a = 5 kph per second

Force required = ma
= 572 x 1.39 = 794.44 N
Force required to accelerate = 794.44 N
The specific heat of the unknown substance with a mass of 0.158kg is 0.5478 J/g°C
HOW TO CALCULATE SPECIFIC HEAT CAPACITY:
The specific heat capacity of a substance can be calculated using the following formula:
Q = m × c × ∆T
Where;
- Q = quantity of heat absorbed (J)
- c = specific heat capacity (4.18 J/g°C)
- m = mass of substance
- ∆T = change in temperature (°C)
According to this question, 2,510.0 J of heat is required to heat the 0.158kg substance from 32.0°C to 61.0°C. The specific heat capacity can be calculated:
2510 = 158 × c × (61°C - 32°C)
2510 = 4582c
c = 2510 ÷ 4582
c = 0.5478 J/g°C
Therefore, the specific heat capacity of the unknown substance that has a mass of 0.158 kg is 0.5478 J/g°C.
Learn more about specific heat capacity at: brainly.com/question/2530523