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2 years ago

What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest

1 answer:

5 0

|acceleration| = (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec

|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.

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Sharon is driving on a straight road. She is driving north, and her speed is

sdas [7]

Answer:

Explanation:

b is the correct answer

4 0

2 years ago

Which electromagnetic wave is used in dental scans?

julsineya [31]

Answer:

X-rays are most often used to examine bones and teeth.

Hope this helps!

8 0

3 years ago

Read 2 more answers

What linear speed must an earth satellite have to be in a circular orbit at an altitude of 159 km?

IgorLugansk [536]

Gravitational acceleration, g = GM/r^2. Additionally, for a satellite in a circular orbit, g = v^2/r

Where, G = Gravitational constant, M = Mass of earth, r = distance from the center of the earth to the satellite, v = linear speed of the satellite.

Equating the two expressions;
v^2/r = GM/r^2
v = Sqrt (GM/r);
But GM = Constant = 398600.5 km^3/sec^2
r = Altitude+Radius of the earth = 159+6371 = 6530 km

Substituting;
v = Sqrt (398600.5/6530) = 7.81 km/sec = 781 m/s

8 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

Oksana_A [137]

Answer:

a.Attractive

$2.31531\times 10^{-16}\ N$

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

$q_1=q_2$ = Charge of electron and proton = $1.6\times 10^{-19}\ C$

r = Distance between them = 997 nm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N$

The force of attraction between the particles will be $2.31531\times 10^{-16}\ N$

8 0

2 years ago

Does anyone know this?!

Valentin [98]

Answer:

2 is the numerical answer.

Explanation:

Hello there!

In this case, according to the given information and formula, it is possible for us to remember that equation for the calculation of the average kinetic energy of a gas is:

$KE=\frac{3}{2} \frac{R}{N_A} T$

Whereas R is the universal gas constant, NA the Avogadro's number and T the temperature.

Which means that for the given ratio, we can obtain the value as follows:

$=\frac{\frac{3}{2} \frac{R}{N_A} T_1}{\frac{3}{2} \frac{R}{N_A} T_2} \\\\=\frac{T_1}{T_2} \\\\=\frac{500K}{250K} \\\\=2$

Regards!

8 0

2 years ago