The engine on a fighter airplane can exert a force

A car traveled at an average speed of 60 mph for two hours. How far did it travel?

Akimi4 [234]

The answer is C- 120 Miles.

6 0

3 years ago

2. A girl whose mass is 55kg stands on a spring weighing machine inside a lift. When the lift starts to ascend, its acceleration

Annette [7]

Answer:

b

Explanation:

5 0

3 years ago

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

Musya8 [376]

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, $M=1.99\times 10^{30}\ kg$

Radius of Mercury's orbit, $r=5.79\times 10^{10}\ m$

Radius of discovered planet, $R=\dfrac{2}{3}r$

$R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m$

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

$T^2\propto R^3$

$T^2=\dfrac{4\pi^2R^3}{GM}$

$T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}$

$T=\sqrt{1.71\times 10^{13}}$

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

6 0

3 years ago

In a car moving at constant acceleration, you travel 230 m between the instants at which the speedometer reads 40 km/h and 70 km

Goryan [66]

The relationship between the distance covered, initial and final speeds, and time can be expressed through the equation,

First equation,

2ad = Vf² - Vi²

Substituting the known values,
2(a)(0.230 km) = (70 km/h)² - (40 km/h)²
The value of a from the equation is 7173.92 km/h².

Second equation,
d = (Vi)(t) + 0.5at²

Substituting the known values,
0.230 km = (40 km/h)(t) + (0.5)(7173.92 km/h²)(t²)

The value of t from the equation is 4.1818 x 10^-3 hours which is also equal to 0.2509 minutes or 15 seconds.

Answer: 15 seconds

7 0

3 years ago

1. The picture below shows Jamal pushing with a 100-Newton (N) force on a large box. Neither

r-ruslan [8.4K]

Answer:

100 newtons

Explanation:

Given,

Jamal pushing a large box by a force, F = 100 N

Work done on the large box is, W = 0

It is because the applied force is less than the force of the friction between the two surfaces.

Yet, there will be a force that is exerted by the large box on Jamal.

According to newton's third law of motion, every action has an equal and opposite reaction. The reaction force is in the direction opposite to the force of action. But, their magnitude remains the same.

$F_{a} =-F_{r}$

Hence, If the action force is 100 N, then the reaction force should be in 100 N

4 0

3 years ago

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