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soldi70 [24.7K]
2 years ago
9

A ball starts from rest and rolls down a hill at a constant acceleration of 5 m/s2. If it travels for 8 m how fast is it going i

n the end?
Physics
1 answer:
makkiz [27]2 years ago
5 0

Hi there!

We can use the following kinematic equation:

v_f^2 = v_i^2 + 2ad

vf = final velocity (? m/s)
vi = intial velocity (0 m/s)

a = acceleration (5 m/s²)

d = displacement (8 m)

Plug in the givens and solve.

v_f^2 = 0 + 2(5)(8)\\\\v_f = \sqrt{80} = \boxed{8.944 \frac{m}{s}}

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A) no H30+ ions or OH- ions.
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Explain why the types of technology valued can vary.
Elan Coil [88]
Over time, the types of technology can vary and be improved upon so that more advanced techniques become more valued. This could be the situation with mining whereby back in the 1500's in underground mines the rock was broken by fire setting ie lighting a fire below the rock face to heat up the rock and then throwing cold water on it to crack it, so that it could be dug by hand. With the advent of explosives, this all changed so that the rock could be blasted. The increase in advance rates for an underground heading have thus gone from 5-20 feet per month to up to 300meters (984 ft) per month for a 24/7 mining operation, which is a huge improvement.
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3 years ago
Read 2 more answers
Calculate the net force on particle q1. First, find the direction of the force particle q2 is exerting on particle q1. Is it pus
ValentinkaMS [17]

The net force on particle particle q1 is 13.06 N towards the left.

<h3>Force on q1 due to q2</h3>

F(12) = kq₁q₂/r₂

F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)

F(12) = -14.41 N  (towards left)

<h3>Force on q1 due to q3</h3>

F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)

F(13) = 1.352 N (towards right)

<h3>Net force on q1</h3>

F(net) = 1.352 N - 14.41 N

F(net) = -13.06 N

Thus, the net force on particle particle q1 is 13.06 N towards the left.

Learn more about force here: brainly.com/question/12970081

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8 0
1 year ago
Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temp
zepelin [54]

Answer: The cylinder

Explanation:

Among all other solid shapes, the sphere has the smallest area for a given volume.

By experiment, the ratio of the radius of a sphere to a cylinder of equal volume is less than 1.

Recall;

That the Rate of transfer of convective heat (Q) = h × A ×change in temperature.

Where ,

h= the co efficient of convective heat transfer

A= the cross sectional area.

As such, since the sphere has a smaller surface area relative to the cylinder, the sphere transfers heat slower than the cylinder.

Therefore, if the sphere and cylinder are exposed to convection in the same environment, then, the cylinder cools faster.

PS; the more the Area, the higher the rate of heat transfer and vice versa.

7 0
3 years ago
n an experiment of a simple pendulum, measurements show that the pendulum has length m, mass kg, and period s. Take m/s2 . i. Us
barxatty [35]

Answer:

The answer is "(1.265 \pm 0.010) \ s \ and \ 0.709 \%"

Explanation:

In point i:

T_{theo}= 2\pi \sqrt{\frac{l}{g}}

        =2\pi\sqrt{\frac{0.397}{9.8}}\\\\= 1.265 \ s

If  error in the theoretical time period :

\frac{\Delta T_{theo}}{T_theo} = \frac{1}{2}  \frac{\Delta l }{l}\\\\\Delta T_{theo} = 1.265 \times \frac{1}{2} \times \frac{0.006}{0.397}

           = 0.010 \ s

 T_{theo} = (1.265 \pm 0.010) \ s

In point ii:

\% \ difference = \frac{|T_{exp} -T_{theo}|}{\frac{T_{exp}+T_{theo}}{2}} \times 100

<h3>                     = \frac{1.274 -1.265}{\frac{1.274+1.265}{2}} \times 100\\\\=0.709 \%</h3>
5 0
3 years ago
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