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mario62 [17]
4 years ago
10

Whwhich pf the fallowing equations can be used to directly calculate an objects impulse

Physics
2 answers:
Bad White [126]4 years ago
8 0
Since I don't see a picture, the impulse is
p=m*v
where m is the mass
and is the velocity
sashaice [31]4 years ago
5 0
P = mass times velocity need to show a pic to get help
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Please help!
o-na [289]
The formula PE(Potential Energy)= mgh

4 0
3 years ago
The moon's gravity is one-sixth that of the earth. what is the period of a 2.00m long pendulum on the moon
erik [133]

Answer:

6.96 s

Explanation:

The period of a simple pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum and g the acceleration due to gravity.

In this problem, we have a pendulum with length L = 2.00 m, while the acceleration due to gravity is 1/6 that of the earth:

g' = \frac{g}{6}=\frac{9.8 m/s^2}{6}=1.63 m/s^2

So, the period of the pendulum on the moon is

T=2 \pi \sqrt{\frac{2.00 m}{1.63 m/s^2}}=6.96 s

3 0
3 years ago
What does the bohr model look like?
sergeinik [125]
Pic of Bohr Model attached

7 0
4 years ago
Read 2 more answers
An atom of uranium 238 emits an alpha particle (an atom of He) and recoils with a velocity of 1.895 * 10^ 5 m/sec . With velocit
lora16 [44]

<u>Answer:</u> The velocity of released alpha particle is 1.127\times 10^7m/s

<u>Explanation:</u>

According to law of conservation of momentum, momentum can neither be created nor be destroyed until and unless, an external force is applied.

For a system:

m_1v_1=m_2v_2

where,

m_1\text{ and }v_1 = Initial mass and velocity

m_2\text{ and }v_2 = Final mass and velocity

We are given:

m_1=238u\\v_1=1.895\times 10^{5}m/s\\m_2=4u\text{ (Mass of }\alpha \text{ -particle)}\\v_2=?m/s

Putting values in above equation, we get:

238\times 1.895\times 10^5=4\times v_2\\\\v_2=\frac{238\times 1.895\times 10^5}{4}=1.127\times 10^7m/s

Hence, the velocity of released alpha particle is 1.127\times 10^7m/s

4 0
3 years ago
A gymnast of mass 70.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
DiKsa [7]

Answer:

T = 686.7N

Explanation:

For this exercise we will use Newton's second law in this case there is no acceleration,

      ∑ F = ma

      T -W = 0

The gymnast's weight is

     W = mg

We clear and calculate the tension

     T = mg

     T = 70 9.81

     T = 686.7N

3 0
3 years ago
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