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3 years ago

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A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box t

o continue sliding. If the student pushes with a constant 10 N force, what is the box's speed when it is released?

1 answer:

Marrrta [24]3 years ago

6 0

Answer:v=3.08 m/s

Explanation:

Given

mass of student $m=21 kg$

distance moved $d=10 m$

Force applied $F=10 N$

acceleration of system during application of force is a

$a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2$

using $v^2-u^2=2 as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-0=2\times 0.476\times 10$

$v=\sqrt{9.52}$

$v=3.08 m/s$

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A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c

Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 $m/ s^{2}$

$F - w = ma \\$

$F = w + ma$

$F = m ( a +g )$

$F = 75 ( 1.5 + 10 ) \\$

$F = 75 ( 11.5 )$

$F = 862.5 N$

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.

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2 years ago

The question states: two large, parallel conducting plates are 12cm

AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, $E=\frac{F}{q}$

$E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}$

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0

3 years ago

PLEASE HELP MEE PLEASE I BEG

TEA [102]

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

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3 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force $F_c$ . In this case, the only force that applies for that is the static frictional force $f_s$ between the tires and the track. Then, we can write that:

$f_s=F_c$

And since $f_s\leq \mu N$ and $F_c=\frac{mv^{2}}{r}$ , we have:

$\mu N\geq \frac{mv^{2}}{r}$

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

$N-mg=0\\\\\\implies N=mg$

Substituting this expression for $N$ and solving for $v$ , we get:

$\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}$

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

$v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s$

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

7 0

3 years ago

Read 2 more answers

19. A person pushes with 6.0 N for 4.0 seconds on a 2.0 kg object.

hram777 [196]

Answer:24NS

Explanation:

Impulse=force x time

Impulse=6 x 4

Impulse=24NS

5 0

3 years ago