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Fudgin [204]
3 years ago
13

An objec with a mass of 7 kg accelerates 5m/s2 when an unknown force is applied to it.what is the magnitude of the unknown force

?
Physics
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

<h2>35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7 × 5

We have the final answer as

<h3>35 N</h3>

Hope this helps you

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia
melisa1 [442]

Answer:

H_2 = 91.55 km

Explanation:

Gravity on the surface of planet is given as

g = \frac{GM}{R^2}

as we know that

M = 8.93 \times 10^{22} kg

R = 1821 km

now gravity on the planet is

g = \frac{(6.67 \times 10^{-11})(8.93 \times 10^{22})}{(1821 \times 10^3)^}

so we have

g = 1.8 m/s^2

now we know that

H_{max} = \frac{v^2}{2g}

so we will say

\frac{H_1}{H_2} = \frac{g_2}{g_1}

\frac{500}{H_2} = \frac{9.81}{1.8}

H_2 = 91.55 km

5 0
3 years ago
Discuss how a photon (aka the light particle) can be affected by gravity despite being massless.
Over [174]

Explanation:

Light is clearly affected by gravity, just think about a black hole, but light supposedly has no mass and gravity only affects objects with mass. On the other hand, if light does have mass then doesn't mass become infinitely larger the closer to the speed of light an object travels.

4 0
3 years ago
In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre
sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is 5.625\times10^{-2}\ m

Explanation:

Given that,

Wave length = 500 nm

Radius d= 40\ \mu m

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

\sin\theta=\dfrac{y}{D}

\sin\theta=\dfrac{y}{6}

Using formula for dark fringe

(n-\dfrac{1}{2})\lambda=2d\sin\theta

Put the value into the formula

(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}

y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}

y=0.05625\ m

y=5.625\times10^{-2}\ m

Hence, The distance of the second dark band away from the central bright spot be located is 5.625\times10^{-2}\ m

6 0
3 years ago
A 450 N trunk rests on a 30 degree inclined plane. What is the force acting down the
Minchanka [31]

Answer: What is the force acting down the plane?

• Opp= hyp (sinq)= 450(sin30)= 225 N

• What is the force acting perpendicular to the plane?

• adj= hyp (cosq)= 450(cos30)= 390 N

Explanation:

4 0
3 years ago
A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia
slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0
3 years ago
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