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3 years ago

If I have a mass of 50 kg that has an acceleration of 10 m/s/s, what is the Force?

Physics

1 answer:

Eddi Din [679]3 years ago

4 0

Answer:

To solve this problem first we need to find out the “ Acceleration=a”. We know that,

a=(v-u)÷t

or, a={(80–10) m/s}÷(14 s)

so, a=5 m/s^2

[ here, v=80 m/s, u=10 m/s, t=14 s, a=Acceleration]

Now,

we know that “Force=mass×acceleration”

or,Force=(50×5) N

so, Force=250 N.

So, the applied force on that object is “250 N”.

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How much heat is needed to raise the temperature of 200 g of lead (c = 0.11 kcal/kg ∙ °c) by 10 c°?

Rufina [12.5K]

Heat = mass (m)*specific heat (C)* change in temperature (Δt)

In the current scenario,
mass = 200 g = 0.2 kg
C = 0.11 kCal/kg.°C
Δt = 10 °C

Therefore,
Heat = 0.2*0.11*10 = 0.22 kCal = 0.22*4186 J = 920.92 J

3 0

3 years ago

Which of the following is an engineering solution

Alona [7]

By "solution" it means a course of action that, once carried out, brings about some desired state of affairs. The use of engineer in this context is as a verb meaning "to arrange or bring about through skillful, artful contrivance."

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4 years ago

Read 2 more answers

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

A caterpillar climbs up a one-meter wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the caterpil

Brut [27]

100cm to go in 600 secs = 10mins

2 positive, 1 negative .... net 1 positive per step

6 0

3 years ago

Read 2 more answers

If2.0J of work is done in raising a 180g apple how far is it lifted?

Alex

We know, W = F * s
W = mg * s

Here, w = 2 J
m = 180 g = 0.180 Kg
g = 9.8 m/s

Substitute their values into the expression:
2 = 0.180*9.8 * s
1.764s = 2
s = 2 / 1.764
s = 1.13 meter

In short, Your Final Answer is 1.13 m

Hope this helps!

7 0

3 years ago