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3 years ago

Calculate the force it would take to accelerate a 250 kg motorcycle at a rate of 5 m/s2

Physics

2 answers:

Sphinxa [80]3 years ago

7 0

Force = mass x acceleration

Mass is given to be 250kg

Acceleration is given to be 5m/s^2

Force = 250kg x 5m/s^2 = 1,250 kgm/s^2 = 1,250 Newtons

evablogger [386]3 years ago

5 0

Answer:

1,250

Explanation:

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Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

beks73 [17]

Answer:

a. the force between them quadruples

Explanation:

The electrostatic force between two charges is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

In this problem, the charges on both objects are doubled, so

$q_1' = 2q_1\\q_2' = 2q_2$

While the distance does not change, so the new force will be

$F'= k \frac{(2q_1)(2q_2)}{r^2}=4 (k\frac{q_1 q_2}{r^2})=4 F$

so, the force will quadruple.

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3 years ago

This is the si unit of current

skelet666 [1.2K]

Answer:

ampere

and it is denoted by A

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2 years ago

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A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=880mmHg\\T_1=20^0C=(20+273)K=293K\\P_2=1050mmHg\\T_2=?$

Putting values in above equation, we get:

$\frac{880mmHg}{293K}=\frac{1050mmHg}{T_2}\\\\T_2=350K$

Hence, the final temperature is 350 K

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3 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

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3 years ago

If you jumped out of a plane, you would begin speeding up as you fall downward. Eventually, due to wind resistance, your velocit

MrRa [10]

Answer:

Mg or your weight.

Explanation:

When your velocity is constant, the net force acting on you is 0. That means the upwards force of air resistance must fully balance the downwards force of gravity on you, which is Mg.

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3 years ago