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3 years ago

What is the mass of a car that has a kinetic energy if 4,320,000 J moving at 23 m/s?

Physics

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

kinetic energy=1/2mv^2.

which is 4320000=1/2×m×23^2.

which is 4320000=1/2×m×529.

4320000=264.5m.

m=4320000/264.5.

m=16332.70~16333g

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If Earth was with no tilt, would we still have seasons at all? If so, how would they be different?

murzikaleks [220]

Answer:

If earth had no tilt, we would have no seasons.

Explanation:

As stated in the answer, if the earth had no tilt we wouldn't have seasons. The earth all around the globe would maintain the same temperature,

And due to the no tilt it would also change our orbit to a bit larger slant, in January when we are at our closest to the sun we WOULD have a mini summer. For the North and South Pole, they would remain cold.

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3 years ago

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3 years ago

Read 2 more answers

Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiem

statuscvo [17]

Answer:

t = 0.33h = 1200s

x = 18.33 km

Explanation:

If the origin of coordinates is at the second car, you can write the following equations for both cars:

car 1:

$x=x_o+v_1t$ (1)

xo = 10 km

v1 = 55km/h

car 2:

$x'=v_2t$ (2)

v2 = 85km/h

For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:

$x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}$

$t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s$

The position in which both cars coincides is:

$x=(55km/h)(0.33h)=18.33km$

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2 years ago

A constant force of 45 N directed at angle θ to the horizontal pulls a crate of weight 100 N from one end of a room to another a

Rasek [7]

Answer:

$W=173.48J$

Explanation:

information we know:

Total force: $F=45N$

Weight: $w=100N$

distance: $4m$

vertical component of the force: $F_{y}=12N$

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In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: $F_{x}=Fcos\theta$