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3 years ago

Calculate the kinetic energy of an 800 kg car when it is going at

Physics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

I A 1000 kg car is traveling at 20 m/s. How much kinetic energy does it. Here, Given- mass=1000kg Speed=20m/s Formula ...

Missing: 13.4m/s ‎| Must include: 13.4m/s

Explanation:

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Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts hi

Sati [7]

Answer:

a) v₀ = - 164.62 m / s , c) y = 122.5 m

Explanation:

We can solve this exercise using the free fall kinematic relations.

We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m

For the boy

y = y₀ + v₀ t - ½ g t²

with free fall its initial speed is zero

y = ½ g t2

For superman

y = y₀ + v₀ (t-5) - ½ g (t-5)²

how superman grabs the lot just before hitting the ground

we look for the time it takes the boy down

t = √ (2 y₀ / g)

t = √ (2 180 / 9,8)

t = 6.06 s

in the equation for superman, we clear the volume and calculate

v₀ (t-5) = -y₀ + ½ g (t-5)²

v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²

v₀ 1.06 = -174.49

v₀ = - 174.49 / 1.06

v₀ = - 164.62 m / s

the negative sign indicates that the initial speed is down

b) to graph the position of the two we use the table

t (s) Y_boy (m) Y_superman (m)

0 180 180

1 175.1 180

5 57.5 180

6 3.6 10.18

see attachment for the two curves

c) calculate the height that falls a lot in the 5 seconds (t = 5)

y = -1/2 g t²

y = ½ 9.8 5²

y = 122.5 m

for this height superman has not yet left the skyscraper, so the boy hits the ground

8 0

3 years ago

If the mass of a material is 122 grams and the volume of the material is 10 cm3, what would the density of the material be?

sergij07 [2.7K]

Since d=m/v. 122 g (mass) divided by 10 cm3 (volume)=12.2 g/cm3
Your answer would be 12.2 g/cm3

3 0

4 years ago

A particle initially located at the origin has an acceleration of a=3jm/s^2 and an initial velocity of Vi=5im/s.

-BARSIC- [3]

Given the particle's acceleration is

$\vec a(t) = \left(3\dfrac{\rm m}{\mathrm s^2}\right)\vec\jmath$

with initial velocity

$\vec v(0) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath$

and starting at the origin, so that

$\vec r(0) = \vec 0$

you can compute the velocity and position functions by applying the fundamental theorem of calculus:

$\vec v(t) = \vec v(0) + \displaystyle \int_0^t \vec a(u)\,\mathrm du$

$\vec r(t) = \vec r(0) + \displaystyle \int_0^t \vec v(u)\,\mathrm du$

We have

• velocity at time t :

$\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \displaystyle \int_0^t \left(3\dfrac{\rm m}{\mathrm s^2}\right)\,\vec\jmath\,\mathrm du \\\\ \vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath \\\\ \boxed{\vec v(t) = \left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath}$

• position at time t :

$\vec r(t) = \displaystyle \int_0^t \left(\left(5\dfrac{\rm m}{\rm s}\right)\,\vec\imath + \left(3\dfrac{\rm m}{\mathrm s^2}\right)u\,\vec\jmath\right) \,\mathrm du \\\\ \boxed{\vec r(t) = \left(5\dfrac{\rm m}{\rm s}\right)t\,\vec\imath + \frac12 \left(3\frac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath}$

8 0

3 years ago

What is the magnitude of the net force needed to bring a 2110 kg car to rest from 18.1 m/s in 7.0 seconds?

lozanna [386]

Here is what we know:
a = ?, s = ?, u = 18.1m/s, v = zero/rest, t = 7.0s, m = 2110kg

(a = acceleration, s= displacement, u = initial velocity, v = final velocity, t = time and m is mass)

Now we choose a kinematic formula. Since we know v, u and t, we will use the formula: v = u+at and rearrange it so that we can find a.
a = v-u/t
a = 0-18.1/7.0 = -2.5857...
therefore, a = -2.6m/s

We have our acceleration, now let’s find the net force. To find the force we use one of Newtons laws of motion.
We will use Newtons second law since it describes what happens when one or more forces act upon an object.
F = ma
F = (2110kg)(-2.6m/s)
F = -5486 kg
Therefore the net force F = -5486 N

4 0

4 years ago

5. What is the force of gravity on a

zvonat [6]

Answer:

32.40 N

Explanation:

Acceleration due to gravity on the moon is 1.62 m/s².

W = mg

W = (20 kg) (1.62 m/s²)

W = 32.40 N

3 0

4 years ago