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zhannawk [14.2K]
3 years ago
11

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object?

Physics
1 answer:
hichkok12 [17]3 years ago
7 0

Hello!

A force of 5 N produces an acceleration of 2 m/s2 on the object. What is the mass of the object ?

Data:

F (force) = 5 N

m (mass) = ?

a (acceleration) = 2 m/s²

Solving:

F = m*a

5 = m*2

2\:m = 5

m = \dfrac{5}{2}

\boxed{\boxed{m = 2.5\:kg}}\end{array}}\qquad\quad\checkmark

Answer:  

2.5 kg  

_______________________________  

I Hope this helps, greetings ... Dexteright02! =)

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A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

Answer:

15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

acceleration, a =5 m/s^2

frictional force, F_f =?

net force = ma

net force = F_h - F_f = 40N - F_f

40  - F_f = 5 x 5

- F_f = 25 - 40

multiply both side by -1

F_f = 40 - 25 = 15

the frictional force is 15N

4 0
3 years ago
An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
Serjik [45]

The image of the object is 8cm to the left of the lens (D)

<h3></h3>

What is the image of an object?

The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.

It is calculated thus:

1÷v = 1÷f - 1÷u

<h3>How to calculate the image of an object</h3>

From the formula

1÷v = 1÷f - 1÷u

<h3>Where </h3>

V = image distance fromthe object

U = object

f = focal length

Substitute the values

1÷v = 1÷8 - 1÷ 4

1÷v = - 1÷8

Make v the subject of formula

v = -8cm

Therefore, the image of the object is 8cm to the left of the lens (D)

Learn more on focal length here:

brainly.com/question/25779311

#SPJ1

6 0
2 years ago
An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -
nordsb [41]
<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

                 E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

4 0
3 years ago
PLEASE HELP WILL MARK BRAINLIEST
aniked [119]
2kg objects in the world of a hunting group
6 0
2 years ago
Two 2 kg masses is placed at either end of a rod that has a mass of .5 kg and a length of 3 m. What is the moment of inertia if
sergij07 [2.7K]

Explanation:

a) I=\displaystyle \sum_{i}m_ir_i^2

where r_i is the distance of the mass m_i from the axis of rotation. When the axis of rotation is placed at the end of the rod, the moment of inertia is due only to one mass. Therefore,

I= mr^2 = (2\:kg)(3\:m)^2 = 18\:kg-m^2

b) When the axis of rotation is placed on the center of the rod, the moment is due to both masses and the radius r is 1.5 m. Therefore,

\displaystyle I= \sum_{i}m_ir_i^2 = 2(2\:kg)(1.5\:m)^2 = 9\:kg-m^2

7 0
3 years ago
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