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sp2606 [1]
3 years ago
12

How much force must be applied to a 4 kg to get it to accelerate at 3 m/s^2?

Physics
1 answer:
labwork [276]3 years ago
6 0

Answer:

12 g i think

Explanation:

brainliest pls

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if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?
laiz [17]

Answer:

The 39.

Explanation:

8 0
3 years ago
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Two​ high-speed ferries leave at the same time from a city to go to the same island. The first​ ferry, the​ Cat, travels a
kumpel [21]

Answer:

1 hour

Explanation:

Since the definition of velocity is v=x/t, we can calculate the position at time t of an object moving at speed v with x=vt.

The position of the first ferry is x_1=v_1t

The position of the second ferry is x_2=v_2t

We want to know when they will be 6 miles apart. This means, if we call that distance d, that we want to know when the difference between their positions will be d, or x_1-x_2=d (we know that at the beginning the position of the ferry 1 is of higher value than that of ferry 2 since it left before).

We use our previous formulas then:

d=x_1-x_2=v_1t-v_2t=(v_1-v_2)t

Since we want the time, we do:

t=\frac{d}{v_1-v_2}

And substitute our values:

t=\frac{6\ miles}{30\ miles/hour-24\ miles/hour}=\frac{6\ miles}{6\ miles/hour} =1\ hour

7 0
3 years ago
Bob travels 36m in 40seconds. How fast does he go?
mylen [45]

Answer:

<h2>0.9 Meters Per Second</h2><h2>54 Meters Per Minute</h2><h2>3.24 Kilometers Per Hour</h2>

Explanation:

Hope this helps! <3

( Pic Below )

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3 years ago
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If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude
Mrrafil [7]

The magnitude of the electric field at this location is 1.5\times 10^{10} N/C

<u>Explanation:</u>

Given

Charge\ of\ the\ particle\ Q=4.3\times 10^-18\ C\\Force\ with\ which\ it\ is\ attracted\ F=6.5\times 10^-8\ N

Electric field at this location determined by the force and charge.

E=F/Q

E=\frac{6.5\times 10^-8}{4.3\times10^-18} =1.5\times\ 10^{10}  \ N/C

7 0
4 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
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