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3 years ago

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What mathematical formula correctly expresses the relationship between force, mass and acceleration? A. m = F + a B. F = m / a C

. a = mF D. F = ma

1 answer:

IRINA_888 [86]3 years ago

3 0

D. F=ma

F is for force, and that equals two things, M for mass and A for acceleration. When mass is accelerated, it gives you force. Force equals multiplying mass and its acceleration.

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The volumes of the block is 10,500 mL and density is 8.52 g/mL. Find the block's mass Don't forget units and a space

Schach [20]

Answer:

<h2>89,460 g</h2>

Explanation:

The mass of a substance when given the density and volume can be found by using the formula

mass = Density × volume

From the question we have

mass = 8.52 × 10,500

We have the final answer as

<h3>89,460 g</h3>

Hope this helps you

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3 years ago

If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: $P*V=n*R*T$

where P is the gas' pressure, V its volume, n the number of moles of gas, R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

$P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450$

So the pressure must have gone up to 450 kilopascals.

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3 years ago

Remove the ice cube tray and the plastic bag from the freezer and let them sit for 3 minutes. Then gently remove the piece of ic

Salsk061 [2.6K]

Answer:

That when you leave the ice setting without foil, it melts slower. When you put the piece of ice in the foil it melts faster than just letting it set without foil. I really hope this helps.

Explanation:

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3 years ago

A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

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7 0

3 years ago

A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo

guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}$

If we consider than the direction of the electric field is $\vec{E}=E_0\hat{x}$ , we can solve the problem differentiating the integral for each face of the cube:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6$

E₀ is a constant and each surface is equal to each other, so: $S_1=S_2=S_i=S$

Therefore:

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0$

$\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c$

3 0

3 years ago