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mestny [16]
3 years ago
6

2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?

Physics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

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The equation for free fall at the surface of a celestial body in outer space​ (s in​ meters, t in​ seconds) is sequals10.04tsqua
Sindrei [870]

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

s=\frac{1}{2}at^2

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

s=10.04 t^2

this means that

\frac{1}{2}g = 10.04

so the acceleration of gravity on the body is

g=2\cdot 10.04 = 20.08 m/s^2

The velocity of an object in free fall starting from rest is given by

v=gt

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s

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When a child does not know how to react emotionally to a situation they will observe the emotional expressions of others. This i
Otrada [13]

Answer:

4)

social referencing

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Practice with Density
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Answer:

5.7 g/cm3

Explanation:

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A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
From a cliff 37.6 m high. At the level of the sea, a rock sticks out a horizontal distance of 12.12 m. The acceleration of gravi
Aleks [24]

Answer:

4.3 m/sec

Explanation:

Here height of cliff = y = 37.6 m

Gravitational acceleration = g = 9.8 m/sec2

vi = 0 m/s

Let's find the time which the diver will take if jumps from there!

Using formula

y = vit+1/2gt2

==> 37.6= 0 + 0.5 ×9.8×t^{2}

==>t^{2}= \frac{37.6}{4.9}

==> t = 2.8 sec

In this time the diver has to cover a horizontal distance of 12.12 m

If x = 12.12 m is the horizontal distance to be covered then using

x= Vx × t

==> Vx = x/t

==> Vx= 12.12/2.8 = 4.3 m/s

8 0
3 years ago
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