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3 years ago

8

Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 107 meters from the center of Earth if the force on

a 2.5–kilogram box at the same point is 6.1 newtons.

1 answer:

Sati [7]3 years ago

5 0

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

F = G m₁ m₂ / R² .

If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:

(6.1 N) / (2.5 kg) = (F) / (1 kg)

Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)

Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .

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A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g

mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

$n_{r}=1.572$

$n_{b}=1.587$

We need to calculate the angle for red wavelength

Using Snell's law,

$n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}$

Put the value into the formula

$1.572\sin37=1\times\sin\theta_{r}$

$\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})$

$\theta_{r}=71.0^{\circ}$

We need to calculate the angle for blue wavelength

Using Snell's law,

$n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}$

Put the value into the formula

$1.587\sin37=1\times\sin\theta_{b}$

$\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})$

$\theta_{b}=72.7^{\circ}$

We need to calculate the angle between the red and blue light

Using formula of angle

$\Delta \theta=\theta_{b}-\theta_{r}$

Put the value into the formula

$\Delta \theta=72.7-71.0$

$\Delta \theta=1.7^{\circ}$

Hence, The angle between the red and blue light is 1.7°.

8 0

3 years ago

Each value in nature has a number part, called its____<br> and a dimension, or unit

saul85 [17]

Answer:

<u>Magnitude</u>

Explanation:

Each value in nature has a number part, called its magnitude and a dimension called its unit.

For example,

The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.

7 0

3 years ago

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$

$s_2 = v_ht_2 = 15*3.49 = 52.5m$

8 0

3 years ago

Why is it important to use placebos and a double-blind approach in some studies

viktelen [127]

Another name for these two words is "constant" and you want to have a "constant", because you want something to compare your experimental group to, to see whether data had changed or not. So you have placebos or a double- blind to compare your experimental group to it and also so you know you don't have a bias or anything in the study.

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3 years ago

A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of

PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box. There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0

3 years ago