A car drives on a highway with speed of 68mi/h. what is this speed in km/h?

How much work is done by the force lifting a

Ann [662]

The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

$W=\Delta U=mg \Delta h$

where

m=0.1 kg is the mass of the hamburger

$g=9.81 m/s^2$ is the gravitational acceleration

$\Delta h=0.3 m$ is the increase in height of the hamburger

Substituting numbers into the equation, we find

$W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J$

So, the correct answer is

(3) 0.3 J

3 0

3 years ago

the pull of gravity on mars is 3.7m/s^2. if a astronaut on mars lifts a 10 kg rock 1 m off the ground, just to see whats under i

Elanso [62]

Gravitational potential energy can be calculated using the formula:

$PE_{grav} =mgh$

Where:
PEgrav = Gravitational potential energy
m= mass
g = acceleration due to gravity
h = height

On Earth acceleration due to gravity is a constant 9.8 but since the scenario is on Mars, the pull of gravity is different. In this case, it is 3.7, so we will use that for g.

So put in what you know and solve for what you don't know.
m = 10kg
g = 3.7m/s^2
h = 1m

So we put that in and solve it.
$PE_{grav} =mgh$
$PE_{grav} =(10kg)(3.7m/s^{2})(1m)$
$PE_{grav} =37J$

7 0

3 years ago

Sallys physical education teacher timed her run and recorded the time and distance in the table below. What is her average speed

Fudgin [204]

Answer: B) 2.5 m/s

Explanation: Find the average of the time and distance, and see how far they go in only 1 second.

1 + 2 + 3 + 4 + 5 = 15
15 divided by 5 = 3

3 seconds

2 + 5 + 7 + 10 + 12 = 36
36 divided by 5 = 7.2

7.2m per 3 seconds.

7.2 divided by 3 = 2.4

Therefore, the answer is technically 2.4m/s

4 0

3 years ago

In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu

Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

Speed of proton = $2.50 \times 10^{6} m/s$

Distance traveled = $5.31 \times 10^{-13} m$

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

4 0

3 years ago

The electric potential in a region of space is \[V=350/\sqrt{x ^{2}+y ^{2}}\] where x and y are in meters. what is the strength

VARVARA [1.3K]

So the given value or the formula in getting the electric potential region of space is V=350/sqrt of x^2+y^2. So the given data is x and y is equals to 2.6 and 2.8. So in my calculation i came up with an answer of 91.6

8 0

3 years ago