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1 year ago

A 5.5 kg box is pushed across the lunch table. The net force applied to the box is 9.7N. What is the acceleration of the box?

Physics

1 answer:

12345 [234]1 year ago

8 0

Answer:

1.76 m/s^2

Explanation:

mass of box, m = 5.5kg

net force , F = 9.7N

acceleration = ?

F = m x a

a = F/m = 9.7/5.5

a = 1.76 m/s^2

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The witch in "Hansel and Gretel" owns a house made of cookies and candy for the ___ reason that she is sweet and kind, but actua

hichkok12 [17]

The answer would be A. Ostensible

8 0

1 year ago

7. A girl pushes her little brother on his sled with a force of 300. N for 750. m. How much work is this if the force of frictio

victus00 [196]

Answer:

a) 75000Joules

b) 0Joules

Explanation:

Workdone = Force * Distance

Given

distance= 750m

Force = 300N

a) If the frictional force = 200N

The Total force = 300N - 200N = 100N

Work done = 100 * 750

Workdone = 75,000Joules

Hence the workdone if the force of friction is 200N is 75,000Joules

b) If the frictional force = 300N

The Total force = 300N - 300N = 0N

Work done = 0* 750

Workdone = 0Joules

Hence the workdone if the force of friction is 300N is 0Joules i.e no work will be done on the sled

3 0

1 year ago

A snowmobile has an initial velocity of +3.5 m/s.

KATRIN_1 [288]

The snowmobile's velocity at time t is

v = 3.5 m/s + (-0.89 m/s²) t

We want to know the time t it takes for v to reach 0, so we solve

0 = 3.5 m/s + (-0.89 m/s²) t

3.5 m/s = (0.89 m/s²) t

t = (3.5 m/s) / (0.89 m/s²)

t ≈ 3.93 s

4 0

1 year ago

Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and c

rodikova [14]

Answer:

Pressure of woman will be $99.87\times 10^5N/m^2$

Pressure of the elephant will be $1716560.50N/m^2$

Explanation:

We have given that mass of the woman m = 80 kg

Acceleration due to gravity $g=9.8m/sec^2$

Diameter of shoes = 1 cm =0.01 m

So radius $r=\frac{d}{2}=\frac{0.01}{2}=0.005m$

So area $A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2$

We know that force is given F = mg

So $F=80\times 9.8=784N$

Now we know that pressure is given by $P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2$

Now mass of elephant m = 5500 kg

So force of elephant = 5500×9.8 = 53900 N

Diameter = 20 cm

So radius r = 10 cm

So area will be $A=3.14\times 0.1^2=0.0314m^2$

So pressure will be $P=\frac{53900}{0.0314}=1716560.50N/m^2$

3 0

1 year ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

1 year ago