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3 years ago

Which hand is negatively charged? A B C D

Physics

2 answers:

vichka [17]3 years ago

6 0

Answer:

the one that has more negative sights so the first hand on your right :)

VLD [36.1K]3 years ago

4 0

Answer:

Explanation:

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What is the mass of a football in free fall with a total force due to gravity of 782N?

Mama L [17]

Explanation:

hbmhbhjvjhvkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

4 0

3 years ago

Read 2 more answers

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

5 0

3 years ago

A 0.80 kg basketball traveling upward at 5.0 m/s impacts an 8.0 10 kg tennis ball traveling downward at 5.0 m/s. The basketball’

vlabodo [156]

To solve this problem we will apply the concepts related to the conservation of momentum. This can be defined as the product between the mass and the velocity of each object, and by conservation it will be understood that the amount of the initial momentum is equal to the amount of the final momentum. By the law of conservation of momentum,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_1$ = Mass of Basketball

$m_2$ = Mass of Tennis ball

$u_1$ = Initial velocity of Basketball

$u_2$ = Initial Velocity of Tennis ball

$v_1$ = Final velocity of Basketball

$v_2$ = Final velocity of the tennis ball

Replacing,

$(0.8)(0.5)+(0.1)(-5.0)=(0.8)(0.3)+(0.1)v_2$

Solving for the final velocity of the tennis ball

$v_2 = 11m/s^2$

Therefore the velocity of the tennis ball after collision is 11 m/s

3 0

3 years ago

Read 2 more answers

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

If you wanted to live where the chances of a destructive earthquake were small, would you pick a location near a fault zone, nea

exis [7]

Answer: A volcanic island arc like Hawaii.

Explanation: The Hawaiian island is the most safest place for the people to live in comparison to the other given options.

The plate tectonic movement generates energy due to collision, resulting in sudden release of energy, and seismic waves are produced that propagates and causes earthquake.

Due to the constant collision between the plates, different places at different time experiences earthquake. Near a fault region, earthquakes are very severe, and also near the mid oceanic ridge it is not possible for a person to live. The subduction zones are the region where a denser plate subducts beneath the other, and its impossible for life to exist there.

Thus the most safest place out of all the option is the volcanic islands where numerous people resides. for example, the Hawaiian and Stromboli.

5 0

3 years ago