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EastWind [94]
3 years ago
8

HELP ME PLEASE URGENT WILL MARK AND 5 STARS

Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

option (B) is the correct option.

please thanks me and follow me

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What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?
kipiarov [429]

Answer:

\boxed {\boxed {\sf a= -7 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over time.

a= \frac {v_f-v_i}{t}

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s

Substitute the values into the formula.

a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}

Solve the numerator.

a= \frac { -35 \ m/s}{5 \ s}

Divide.

a= -7 \ m/s/s

a= -7 \ m/s^2

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0
2 years ago
What is the name of the particle found in the shell
elena-s [515]
If you’re talking about atoms, I hope this helps
Atoms consist of three basic particles: protons, electrons, and neutrons. The nucleus (center) of the atom contains the protons (positively charged) and the neutrons (no charge). The outermost regions of the atom are called electron shells and contain the electrons (negatively charged)
7 0
2 years ago
Read 2 more answers
A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity
Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s

With which we compute the maximum height:

y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0
3 years ago
The transformer is based on:
gizmo_the_mogwai [7]

the answer is induction

4 0
3 years ago
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}

With ε0 = 8.85*10^-12 C^2/(N*m^2)

Then:

E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}

E(0.89) = 306500 N/C

3 0
3 years ago
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