All categories

2 years ago

How much force is required to accelerate a 12 kg mass at 5 m/s 2

Physics

1 answer:

Savatey [412]2 years ago

4 0

Answer:

60 N

Explanation:

This is just Newton's Second Law

F = m*a

F = ?

m = 12 kg

a = 5 m/^2

F = 5*12 = 60 Newtons

You might be interested in

In a shipping yard, a crane operator attaches a cable to a 1240-kg shipping container and then uses the crane to lift the contai

Firdavs [7]

Answer:

Explanation:

Given

mass of ship =1240 kg

height=37 m

(a)Work done by Tension in cable+work done by gravity =0

Work done by gravity =change in potential energy of mass $=-1240\times 9.8\times 37=-449.624 kJ$

Thus work done by tension=449.624 kJ

(b)work done by force of gravity $=-mg\times h=-1240\times 9.8\times 37$ =-449.624 kJ

4 0

3 years ago

Read 2 more answers

Two samples of water are mixed together.

Anon25 [30]

<h3><u>Answer;</u></h3>

A.75°C

<h3><u>Explanation</u>;</h3>

Let the change in temp of cold water be x degrees,

while that of hot water be 100 - x degrees.

Heat exchange = mcΔt

Ice

Δt = x

m = 0.50 kg

c = 4.18 kJ/kg*°C

Hot water

Δt = 100 - x

m = 1.5 kg

c = 4.18

But;

Heat lost = heat gained

0.50 * c * x = 1.5 * c * (100 - x)

0.50 *x = 1.5*(100 - x)

0.5x = 150 - 1.5x

0.5x + 1.5x = 150 - 1.5x + 1.5x

2x = 150

x = <u>75° C</u>

Hence; the equilbrium temperature will be 75° C

7 0

3 years ago

What is the most likely reason that Mendeleev placed tellurium before iodine?

Ilya [14]

Mendeleev watched that tellurium has compound properties like different components in its gathering, and he didn't realize that neutrons cause the more noteworthy nuclear mass. Mendeleev expressed that he anticipated that tellurium would have a lower nuclear mass than iodine.

8 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$