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Sidana [21]
3 years ago
13

NEED HELP PLEASE 15 POINTS PLEASEE NEED HELP

Physics
2 answers:
Ostrovityanka [42]3 years ago
5 0

Answer:

I wish I knew

Explanation:

Reil [10]3 years ago
5 0

Answer: The third one

Explanation:

You might be interested in
M = 15 kg, a = 2 m/s2, F =
Lisa [10]

Answer:

<h2>30 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 = 15 × 2

We have the final answer as

<h3>30 N</h3>

Hope this helps you

7 0
3 years ago
The net force (unbalanced force) applied on an object is the product of_________and___________​
RSB [31]

Answer:

of its <u>mass</u> and its <u>acceleration</u>

<u>Explanation:</u>

based on Newton's second law of motion

5 0
1 year ago
If the object represented by the FBD below has a mass of 2.5 kg, what is the acceleration of the object?
Debora [2.8K]

Answer:

4 m/s² down

Explanation:

We'll begin by calculating the net force acting on the object.

The net force acting on the object from the left and right side is zero because the same force is applied on both sides.

Next, we shall determine the net force acting on the object from the up and down side. This can be obtained as follow:

Force up (Fᵤ) = 15 N

Force down (Fₔ) = 25 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fᵤ

Fₙ = 25 – 15

Fₙ = 10 N down

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Mass (ml= 2.5 Kg

Net force (Fₙ) = 10 N down

Acceleration (a) =?

Fₙ = ma

10 = 2.5 × a

Divide both side by 2.5

a = 10 / 2.5

a = 4 m/s² down

Therefore, the acceleration of the object is 4 m/s² down

6 0
3 years ago
Calculate the pressure exerted on the floor by the boy standing on both feet if the weight of the boy is 40kg. Assume that the a
Jlenok [28]

Answer:

P = 1333.33 N

Explanation:

The pressure exerted by the boy on the floor can be calculated by the following equation:

P = \frac{F}{A}

where,

P = Pressure exerted by the boy = ?

F = Force Applied = Weight of Boy = 40 kg = 40 N (since 1 kg = 1N)

A = Area of application of force = 2(Area of one show) = 2(6 cm x 25 cm)

A = 2(0.06 m x 0.25 m) = 0.03 m²

Therefore,

P = \frac{40\ N}{0.03\ m^2}\\\\

<u>P = 1333.33 N</u>

4 0
3 years ago
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r
andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature T_1=781 K

Reservoir 2 Temperature T_2=335 K

Let Q is the amount of heat Flows i.e. Q=1477 J

thus change in Entropy is given by \frac{\sum Q}{T}

\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}

\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}

\Delta S=\frac{\sum Q}{T}=-1.891+4.4089

\Delta S=\frac{\sum Q}{T}=2.517 J/K                              

6 0
2 years ago
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