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sashaice [31]
2 years ago
12

motorcycle accelerates from rest to a final velocity of 27 m/s in 9 seconds. What is its acceleration?

Physics
1 answer:
Vedmedyk [2.9K]2 years ago
5 0

Answer:

3 m/s squared

Explanation:

The formula you use is Vf= Vi + at. You rearrange it to a= Vf - Vi/t. The Vf is 27m/s. The Vi is 0m/s and the t is 9s. Cross out Vi since it’s zero and you’re left with a= 27m/s divided by 9s, which equals 3

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An object has a mass of 7 kg and is accelerating at 4 m/s2. How far would the object move if it took 168 J of work to move it?​
KIM [24]

Answer:

6m

Explanation:

use work =force×distance

6 0
2 years ago
Why do scientists conduct experiments ?
kkurt [141]
Scientists conduct experiments in order to prove a theory or a prediction they have or contradict it, so that then they can write down their results to study them.
8 0
3 years ago
A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
stepan [7]

Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

\rho = \frac{m_1}{V_1}

m_1 = \rho V_1 = 8.38 *2*10^{-3}

m_1 = 0.0167 kg

for m_2

m_2 = \rho V_i - V_removed

m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

\frac{p_2}{p_1}=\frac{m_2}{m_1}

p_2 = p_1 * \frac{m_2}{m_1}

p_2 =7*10^{5} * \frac {.0159}{0.0167}

p_2 = 664081 N/m^{2}

8 0
2 years ago
The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol
Lesechka [4]

Answer:

\Delta V = 0.053 A

Explanation:

Electric field in a given region is given by equation

E = \frac{A}{r^4}

as we know the relation between electric field and potential difference is given as

\Delta V = -\int E. dr

so here we have

\Delta V = - \int (\frac{A}{r^4}) .dr

\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}

here we know that

r_1 = 1.71 m  and r_2 = 2.89 m

so we will have

\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})

so we will have

\Delta V = 0.053 A

8 0
2 years ago
If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula
snow_lady [41]

Here we know that

\omega_i = - 8.4 rad/s

\alpha = - 2.8 rad/s^2

t = 1.5 s

now from kinematics we have

\omega_f = \omega_i + \alpha t

now from above all values we have

\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)

\omega_f = -8.4 + (-4.2)

\omega_f = -12.6 rad/s

so final angular speed is -12.6 rad/s

6 0
3 years ago
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