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faltersainse [42]
2 years ago
10

Acceleration is a change in speed or direction over time. In what two ways does the sled accelerate as it descends?

Physics
1 answer:
Anna007 [38]2 years ago
4 0

Answer:

By force and sloppy surface.

Explanation:

By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.

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A car travels 8km in 7 minutes. Find the speed of the car.
AleksandrR [38]

Answer:

42.6083 mi/h

Explanation:

Given: A car travels 8km in 7 minutes.

To find: Find the speed of the car.

Formula: Speed = \frac{Distance}{Time}

Solution: Since the formula for the speed of an object (which is the car) is speed = distance ÷ time, divide the distance (8km) by the time (7min)

Speed = 42.6083 miles per hour

3 0
2 years ago
What fills the void between stars and galaxies.
Lana71 [14]
Answer: The voids between stars in our galaxy can be filled with tenuous clouds of gas and other molecules. ... That material gets "ripped away" from the galaxies by the force of gravity, and often enough it collides with other material.

HOPE IT HELPED:) HAVE A NICE DAY
6 0
3 years ago
can you suggest improvement that can be made towards the design of siphon so that the transfer of liquid is much higher.​
ahrayia [7]

Answer:

A siphon is a tube that makes use of the potential energy of fluid at an elevated level to transfer the fluid to a lower level, due to pressure differences between the inlet and the outlet points of the tube, such that the pressure at the outlet is higher than the pressure at the inlet

The pressure energy is converted into velocity (kinetic) energy, and therefore, in other to increase the flow rate through the tube of a siphon, with constant diameter, the level of the fluid in the container at the inlet (supply) of the siphon is raised higher than the level at the outlet receiving) container or the outlet point of the siphon tube

The larger the difference between the inlet and outlet levels, the faster the transfer of fluid by the siphon

Explanation:

3 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is
Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

I'm taking the K12 Unit Test: Energy - Part 1 right now

7 0
2 years ago
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