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Snezhnost [94]
3 years ago
8

Is it possible for a tennis ball to have more momentum than a cannonball? explain ​

Physics
1 answer:
Olenka [21]3 years ago
8 0
Well you know how it
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Divers in acapulco, mexico, dive headfirst at 8 feet per second from the top of a cliff 87 feet above the pacific ocean. during
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Depends on the wieght of his genitals.
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3 years ago
A particular roller coaster has a mass of 3500 kg, a height of 4.0 m, and a velocity of 12 m/s. What is the kinetic energy? If n
Alenkinab [10]

Explanation:

given

m=3500kg,h=4.0m,v=12m/s

ke=½mv²

½×3500×12²=3500×144÷2

=252,000j or 252j

8 0
2 years ago
CAN SOMEONE HELP ME PLEASE!? After chasing its prey, a cougar leaves skid marks that are 236 m in length. Assuming the cougar sk
malfutka [58]

Answer:

u=36.8m/s

Explanation:

because of the acceleration is a constant acceleration we can use one of the "SUVAT" equations

u^2=v^2-2ā*s. where:

u^2 stands for intial velocity

v^2 stands for final velocity

since the cougar skidded to a complete stop the final velocity is zero.

u^2=v^2-2ā*s

u^2=(0)^2 -2(-2.87 m/s^2)*236 m

u^2=0+5.74m/s^2* 236m

u^2=1354.64m^2/s^2

u=√1354.64m^2/s^2

u=36.8m/s (approximate value)

when ever the acceleration is constant you can use one of the following equation to find the required value.

1. v = u + at. (no s)

2. s= 1/2(u+v)t. (no ā)

3. s=ut + 1/2at^2. ( no v)

4. v^2=u^2 + 2āS. (no t). 5. s= vt - 1/2at^2. (no u)

5 0
3 years ago
What kind of symmetry do you have
Keith_Richards [23]
During the daytime, I have mostly line symmetry.

During the night, I often have almost spherical symmetry.
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3 years ago
Read 2 more answers
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

3 0
3 years ago
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