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Katen [24]
3 years ago
10

An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?

Physics
2 answers:
Anastaziya [24]3 years ago
5 0

Answer:

acceleration  =  4 m/s²

Explanation:

alexdok [17]3 years ago
4 0
Newton's 2nd law of motion says

                   Net force = (mass) x (acceleration)

Plug in the things you know, and you have

                   80 N  =  (20 kg) x (acceleration)

Divide each side by (20 kg) :

                  80N / 20kg  =  acceleration

                   acceleration  =  4 m/s²
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Lady_Fox [76]

Answer:

Explanation:

Based on the wave model of light, physicists predicted that increasing light amplitude would increase the kinetic energy of emitted photoelectrons, while increasing the frequency would increase measured current.

Contrary to the predictions, experiments showed that increasing the light frequency increased the kinetic energy of the photoelectrons, and increasing the light amplitude increased the current.

Based on these findings, Einstein proposed that light behaved like a stream of particles called photons with an energy of \text{E}=h\nuE=hνstart text, E, end text, equals, h, \nu.

The work function, \PhiΦ\Phi, is the minimum amount of energy required to induce photoemission of electrons from a metal surface, and the value of \PhiΦ\Phi depends on the metal.

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3 years ago
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Which list of elements contains only metals?
fredd [130]
C. those are all metals.
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3 years ago
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Joey is riding in an elevator which is accelerating upwards at 2.0 m/s2. The elevator weighs 300.0 kg, and Joey weighs 60.0 kg.
vodka [1.7K]

4200 N  is the tension in the cable that pulls the elevator upwards.

The correct option is A.

<h3>What does tension ?</h3>

Tension is the force that is sent through a rope, thread, or wire whenever two opposing forces pull on it. Along the whole length of the wire, the tensile stress pulls equally on all objects at the ends. Every physical object that comes into contact with that other one exerts force on it.

<h3>Briefing:</h3>

We employ the following formula to determine the cable's tension.

Formula:

T = mg+ma............ Equation 1

Where:

T is the cable's tension.

M = Mass of the elevator and the Joey

Accelerating with a

g = Gravitational acceleration

Considering the query,

Given:

m = (300+60) = 360 kg

a = 2 m/s²

g = 9.8 m/s²

Substitute these values into equation 2

T = (360×9.8)+(360×2)

T = 3528+720

T = 4248 N

T ≈ 4200 to the nearest hundred.

To know more about Tension visit:

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7 0
1 year ago
A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a
VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

r_1 = 45 sin15 \hat i + 45 cos15 \hat j

r_1 = 11.64 \hat i + 43.46\hat j

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

r_2 = 30 cos15\hat i + 30 sin15 \hat j

r_2 = 28.98\hat i + 7.76 \hat j

now the displacement of the boat is given as

d = r_2 - r_1

d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)

d = 17.34 \hat i - 35.7 \hat j

so the magnitude is given as

d = \sqrt{17.34^2 + 35.7^2}

d = 39.7 km

4 0
3 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
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