an object is accelerated from a constant rate of 5.00 m/s squared what will be its speed after 8.00 seconds

An auditorium measures 43.0m × 18.0m × 15.0m. The density of air is 1.20 kg/m3. What is the volume of the room in cubic feet? (D

frez [133]

The question here would be what is the volume of the room. The density of air that is given has no use. We simply multiply the dimensions given of the room to determine the volume.

<span>43.0m × 18.0m × 15.0m = 11610m^3 ( 3.28 ft / 1 m)^3 = 4.09 x 10^5 ft^3</span>

5 0

3 years ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

4 0

1 year ago

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain:

$\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}$

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0

3 years ago

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and

Alexxandr [17]

Answer:

a

$F =326.7 \ N$