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Kaylis [27]
3 years ago
7

What is the mass of an object that requires 150N of force in order to accelerate it at 10m/s 2

Physics
2 answers:
Anna007 [38]3 years ago
5 0

Answer: 15kg

Explanation:

F = ma ; Newtons 2nd of Motion

150 = m(10)

m = 15kg

baherus [9]3 years ago
4 0
Force= Mass • Acceleration
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A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma
Dahasolnce [82]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

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

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{38}{57} =  \frac{2}{3}   \\  = 0.666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

4 0
2 years ago
GIVING BRAINIEST TO FAST ANSWER!!!!
Vlad [161]

Answer:

The second one a part of it heats up Earth's land and water equally.

Explanation:

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4 0
2 years ago
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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
3 years ago
The earth has a mass ME = 5.98 · 1024 kg and the moon has a mass MM = 7.36 · 1022 kg. The distance from the center of the earth
ivann1987 [24]

Answer:

2 x 10^20 N

Explanation:

Me = 5.98 x 10^24 kg

Mm = 7.36 x 10^22 kg

r = 3.82 x 10^5 km = 3.82 x 10^8 m

The gravitational force between earth and moon is

F = G Me x Mm / r^2

F = (6.67 x 10^-11 x  5.98 x 10^24 x 7.36 x 10^22) / (3.82 x 10^8 x 3.82 x 10^8)

F = 2 x 10^20 N

6 0
3 years ago
I will mark brainliest if right :)
kozerog [31]

Answer:

Explanation:

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4 0
3 years ago
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