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aev [14]
3 years ago
5

PLS HELP! A boy drops a ball from an observation tower. The ball hits the ground in 3.0 s. What is the ball's velocity at the ti

me of impact?
Physics
1 answer:
Levart [38]3 years ago
4 0

Answer:

v = u + at

v = 0 + 9.8 × 3

v = 29.4 m/sec

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Carlos pushes a 3 kg box with a force of 9 newtons. The force of friction on the box is 3 newtons in the opposite direction. Wha
Ivenika [448]
The 'net' force acting on the box is (9 - 3) = 6 newtons
in the direction that Carlos is pushing.

Force = (mass) x (acceleration)

6 = (3) x (acceleration)

Divide each side by 3 :

<em>2 m/s² = acceleration</em>
4 0
3 years ago
Read 2 more answers
Which object had more potential energy when it was lifted to a distance of 1000 centimeters? Show your calculation.
RSB [31]

Explanation:

We Know That

POTENTIAL ENERGY= MASS*g*HEIGHT

When the objects are lifted to same height then the object with heavier mass would have the highest potential energy

.

5 0
2 years ago
Martina has a sample of an unknown substance. She measures the substance. It’s mass is 13.5 grams, and it’s volume is 5 cm3 .Whi
Dominik [7]
Cm^3 is same as mL

13.5 g / 5 mL = 2.7 g/mL
look up densities of metals
aluminum has a density of 2.7 g/mL
6 0
3 years ago
A car on the roller coaster begins with 0j of kinetic energy and 12,928j of potential energy and finishes the track with 3,715j
andre [41]

Answer:

9213 J

Explanation:

Change in Kinetic energy = Change in Potential energy

= 12,928J - 3715J

=9213 J

For more assistance: +1 (304) 223-3136

7 0
3 years ago
The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra
ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{l}

\mu=\dfrac{0.44}{2}=0.22\ kg/m

T=4L^2f\mu

T=4\times (1.75)^2\times (27.5)^2 \times 0.22

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0
3 years ago
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