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Ierofanga [76]
2 years ago
6

A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

. The resistance of the snow and mud also applies a force to the car, which has a magnitude of 7340 N and points due south. What is the acceleration of the car?
Physics
1 answer:
sergiy2304 [10]2 years ago
7 0

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

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Answer:

1.a

2.longer

Explanation:

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Jack tries to place magnets on his refrigerator at home, but they won’t stick. What could be the reason?
saul85 [17]
The most probable reason why the magnets won't stick on the refrigerator is that the body of the refrigerator and the magnets have like poles. If both have negative or both have positive poles facing each other, they will repel. In principle, magnets are attracted to opposite poles and like poles repel. 
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2 years ago
A 1500 W radiant heater is constructed to operate at 115 V. (a) What will be the current in the heater? (b) What is the resistan
OlgaM077 [116]

Answer:

a) I = 13.04 A

b)  R = 8.82 ohms

c) 1291.87 kilocalories are generated an hour.

Explanation:

let P be the power of the heater, V be the voltage of the heater, I be the current of the heater, R be the resistance.

a) we know that:

P = I×V

I = P/V

  = (1500)/(115)

  = 13.04 A

Therefore, the current of the heater is 13.04 A

b) we now have voltage and current, according to Ohm's law:

R = V/I

  = (115)/(13.04)

  = 8.82 ohms

Therefore, the resistance of the heating coil is 8.82 ohms.

c) the number of kilocalories generated in one hour by the heater is just the energy the heater produces in one hour which is given by:

E = P×t

  = (1500)(1×60×60)

  = 5400000 J

since 1 calorie = 4.81 J

1 kilocalorie = 0.001 calories

E = 5400000/4.18 ≈ 1291866.029 calories ≈1291.87 kilocalories

Therefore, 1291.87 kilocalories are produced/generated in one hour.

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2 years ago
Question 2 (1 point)
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Answer:

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3p - 3e = 0 and that leaves 2 neutrons so it will be neutral

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Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed
olga55 [171]

Initially, the energies are:

U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}

At final point, the energies are:

U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0

Using conservation law of energy,

-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}

The equation is further simplified as:

r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}

7 0
1 year ago
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