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lilavasa [31]
2 years ago
8

A car speeds up from 13 m/s to 23 m/s in 30 seconds. What is the acceleration of the car?

Physics
1 answer:
ra1l [238]2 years ago
7 0
Well, <span>v = u + a×t is the equation.</span>
<span>
v: final velocity, which is 23 m/s in this equation.</span>
<span>u: initialo velocity = 13 m/s </span>
<span>a: acceleration = ? </span>
<span>t: time = 30s 
</span>
Your equation would be...

<span>23 = 13 + a×30 </span>
<span>a = (23 - 13) / 30 </span>
<span>a = 1 / 3 </span>
<span>a = 0.333 m/s</span>
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A boy of mass 80 kg slides down a vertical pole, and a frictional force of 480 N acts on him. What is his acceleration as he sli
olga_2 [115]

Answer:

His acceleration is \overrightarrow{a}=4\frac{m}{s^{2}}

Explanation:

Newton's second law states that acceleration of a body is cause by a net force, the relation between them is:

\sum\overrightarrow{F}=m\overrightarrow{a}

On the boy there're acting two forces, his weight (W) that points downward and the frictional force (f) that points upward (they boy moves downward and friction always is opposite to movement). So \sum\overrightarrow{F}=\overrightarrow{W}+\overrightarrow{f} so (1) is:

\overrightarrow{W}+\overrightarrow{f}=m\overrightarrow{a}

Using the positive direction downward weight and gravitational acceleration(g) are positive and friction force is negative:

W-f=m\overrightarrow{a}, solving for a:

\overrightarrow{a}=\frac{W-f}{m}, weight is mg:

\overrightarrow{a}=\frac{mg-f}{m}=\overrightarrow{f}=\frac{(80kg)(10\frac{m}{s^{2}})-(480)}{80}

\overrightarrow{a}=4\frac{m}{s^{2}}

6 0
2 years ago
A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle
denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/s^{2} = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = \frac{8.04N}{70.2Kg} = 0.1130801680m/s^2.

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0
2 years ago
There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame
slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3

Similarly

V_2=\frac{4}{24}\pi d_2^3

Dividing the above two equations, we get

\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3

∴ The larger one hold 163.296 cm³ of paint

5 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
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