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Aleksandr-060686 [28]
3 years ago
9

How does a fuse work?

Physics
1 answer:
Vilka [71]3 years ago
5 0

Answer:

A. a material burns out when current is excessive

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A car traveling at 23m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is the acceleration?
Lena [83]

The car has an initial velocity v_0 of 23 m/s and a final velocity v of 0 m/s. Recall that for constant acceleration,

v=v_0+at

The car stops in 7 s, so the acceleration is

0\,\dfrac{\mathrm m}{\mathrm s}=23\,\dfrac{\mathrm m}{\mathrm s}+a(7\,\mathrm s)

\implies a\approx-3.29\,\dfrac{\mathrm m}{\mathrm s^2}

7 0
3 years ago
An airplane has a starting velocity of 300m/s. It then accelerates at a rate of 45m/s2 for a time of 10s. What is it's final vel
Olenka [21]
A = (v - u) / t

a = acceleration
v = final velocity
u = initial velocity
t = time

45 = (v - 300) / 10

45 × 10 = v - 300

450 + 300 = v

v = 750 m/s

Hope this helps!

P.S. Let me know if you need an explanation
8 0
3 years ago
A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v
hram777 [196]

Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

<em>F = 600N</em>

<em>Hence the net force acting on the car is 600N.</em>

<em></em>

<em></em>

3 0
3 years ago
Select the appropriate shape for the given volume form
vredina [299]
Fist one is a cylinder
the second, i believe is a sphere
the third is a rectangular prism
and the last is the same as the first, a cylinder
3 0
2 years ago
A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on
SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

\sum F = ma

the resultant of the forces must be zero: \sum F = 0 (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

F, the push applied by the worker

F_f=-\mu mg, the force of friction, with \mu=0.25 being the coefficient of friction, m=30.0 kg being the mass of the crate, and g=9.8 m/s^2. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

W=Fd cos \theta

where

F = 73.5 N is the force

d = 4.5 m is the displacement

\theta=0^{\circ} is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

W=F_f d cos \theta

where

F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N is the magnitude of the force of friction

d = 4.5 m is the displacement

\theta=180^{\circ} is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

W=F_f d cos \theta

We notice that both gravity and normal force are perpendicular to the displacement: therefore, \theta=90^{circ}, and so

cos \theta=0

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0
3 years ago
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