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2 years ago

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help

Physics

1 answer:

kompoz [17]2 years ago

6 0

Answer:

0.5 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

a = 10/20

a = 0.5 m/s²

Therefore, the acceleration of the car is 0.5 m/s².

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An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

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3 years ago

Why is it necessary for cells to be so small

madreJ [45]

Answer:

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Explanation:

That is why the cell needs to be small

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2 years ago

The electric field in a region of space increases from 0 0 to 3450 N/C 3450 N/C in 4.40 s. 4.40 s. What is the magnitude of the

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3 years ago

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

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And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

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$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

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$m=\rho.V$

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3 years ago

Winds that blow from the north and south poles

Gennadij [26K]

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3 years ago

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help ​

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help