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coldgirl [10]
2 years ago
11

A baggage handler at an airport applies a constant horizontal force with magnitude F1 to push a box, of mass m, across a rough h

orizontal surface with a very small constant acceleration a.
The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a.
How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box?
Physics
1 answer:
Vikentia [17]2 years ago
6 0

Answer:

F_2 = 2F_1 - F where F is the friction force that is the same for both boxes

Explanation:

So the same boxes at the same mass m would create a same friction force, let this force be F. The net force in the first and 2nd cases would be

- 1st case: F1 - F

- 2nd case F2 - F

According to Newton 2nd law, this net would make acceleration

- 1st case a_1 = (F_1 - F)/m

- 2nd casea_2 = (F_2 - F)/m

Since a_2 = 2a_1

(F_2 - F)/m = 2(F_1 - F)/m

F_2 - F = 2F_1 - 2F

F_2 = 2F_1 - F

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A 50-kg meteorite moving at 1000 m/s strikes Earth. Assume the velocity is along the line joining Earth's center of mass and the
wel

Answer:

a)  amount of kinetic energy converted to internal energy  = 2.5 x 10 raised to power 7 Joule

b) Kinetic energy gained by the earth = 2.1 x 10-16J

c) All the kinetic energy is converted to internal energy and the energy is further converted to thermal energy hence the reason for the hotness at around where the meteorite strikes.

Explanation:

The detailed steps and appropriate application of the law of conservation of momentum is as shown in the attached file.

3 0
2 years ago
When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance?
grandymaker [24]

Answer:

<h2>1.5 ohms</h2>

Explanation:

Power is expressed as P = V²/R

R = resistance

V = supplied voltage

Given P = 600W and V = 30V

R = V²/P

R = 30²/600

R = 900/600

R = 1.5ohms

magnitude of its resistance is 1.5ohms

3 0
2 years ago
A 0.50 kilogram ball is held at a height of 20 meters. What is the kinetic energy of the ball when it reaches halfway after bein
Margarita [4]
Potential energy at any point is (M G H). On the way down, only H changes. So halfway down, half of the potential energy remains, and the other half has turned to kinetic energy. Half of the (M G H) it had at the tpp is (0.5 x 9.8 x 10) = 49 joules.
6 0
2 years ago
Read 2 more answers
A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0
2 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
2 years ago
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