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shtirl
3 months ago
10

After soccer practice coach Miller goes to the roof of the school to retrieve the event soccer balls the height of the school is

3.5 m a soccer ball which leaves the roof with only a horizontal velocity of 15 m/s in the ignoring air resistance how far is the ball above the ground at 0.32 seconds
Physics
1 answer:
Yuliya22 [3.3K]3 months ago
4 0

Answer:

50.2 cm

Explanation:

We have the following data:

Height, h=3.5 m

Initial horizontal velocity, u_x=15 m/s

Time, t=0.32 s

We need to determine how far the ball is from the ground after 0.32 s.

Initial vertical velocity, u_y=0

s=u_yt+\frac{1}{2}gt^2

Where g=9.8 m/s^

s=0+\frac{1}{2}(9.8)(0.32)^2

s=0.502 m

s=0.502\times 100=50.2 cm

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Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
Ostrovityanka [3204]

Answer:

2.32\times 10^{-11}

Explanation:

The first number is 4.48\times 10^{-8}.

The second number is 5.2\times 10^{-4}.

We must multiply these two numbers together.

4.48\times 10^{-8}\times 5.2\times 10^{-4}=(4.48\times 5.2)\times 10^{(-8-4)}\\\\=23.296\times 10^{-12}

In scientific notation: 2.32\times 10^{-11}

Therefore, this is the solution you are looking for.

8 0
4 months ago
A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote
Keith_Richards [3271]
1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
U=q V
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is q=4.0 \mu C = 4.0 \cdot 10^{-6}C, and the potential energy is U=5.2 \cdot 10^{-3} J, thus we may rearrange the earlier formula to find the electric potential at the tip:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V

2) Using this same formula, if the charge changes to q=2.0 \mu C = 2.0 \cdot 10^{-6} C, the resulting electric potential will be:
V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V
8 0
3 months ago
Read 2 more answers
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