Please help on this one

How high would you have to lift a 1000kg car to give it a potential energy of:

Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

For 1.0kJ = 1 x 10³J;

h = $\frac{1000}{9.8 x 1000}$ = 0.102m

5 0

3 years ago

According to a college survey, 22% of all students work full time. find the mean for the number of 3) students who work full tim

attashe74 [19]

Assuming that the students worldwide are being considered, because of the extremely large population, this can be considered as a binomial distribution. A normal distribution is used most usually as a fair approximation of the binomial. The mean is the expectation, therefore:
E[x] = np = (16)(0.22) = 3.52
μ = 3.52

3 0

3 years ago

What do you want to learn about simple machines? (could someone come up with a question about simple machines for me)

bezimeni [28]

Answer:

what are simple machines lol

Explanation:

6 0

3 years ago

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Andrej [43]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

3 0

3 years ago

You drive a car east on the highway at 26 m/s. Another car passes you moving east traveling at 32 m/s. How fast do you view the

tamaranim1 [39]

The speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

<h3>Relative velocity of the car</h3>

The speed of the car passing you is determined by applying relative velocity principle as shown below;

Vr = Va - Vb

Vr = 26 m/s - 32 m/s

Vr = -6 m/s

Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

Learn more about relative velocity here: brainly.com/question/17228388

#SPJ1

3 0

2 years ago