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3 years ago

What is the kinetic energy of a jogger with a mass of 65 kg traveling at a velocity of 2.5 m/s?

Physics

2 answers:

scZoUnD [109]3 years ago

7 0

KE=203.125 J !!!!!!!!!

nekit [7.7K]3 years ago

6 0

Kinetic energy = 1/2mv2

1/2 x 65 x 2.5^2
=203.125J

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A planet has a mass of 5.68 x 1026 kg and a radius of 6.03 x 107 m. What is the weight of a 65.0 kg person on the surface of thi

Paul [167]

To solve this problem, we must know the gravitational force of the planet. The equation would be,

$F=G \frac{ m_{1} m_{2} }{ r^{2} }$

This would calculate the force between two objects with masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2 kg-1 and with r as the distance between the objects.

Thus,

F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65 kg) * ((1/6.03 x 10^7 m)^2)

F = 678 kg/s^2 or 678 N

Answer is letter B.

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3 years ago

1. A 46000N helicopter feels a net force of 9200N. What litt force is exerted by the air on the propellers?

Lina20 [59]

Answer:

Force exerted by the air on the propellers = 46000 - 9200

= 36800 N

Hope this helps!

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3 years ago

How many hydrogen atoms are there in all the reactants?

Alinara [238K]

Answer:

There are two hydrogen atom in all the reactants.

6 0

3 years ago

Read 2 more answers

A 15.0 kg turntable with a radius of 25 cm is covered with a uniform layer of dry ice that has a mass of 9.0 kg. The angular spe

liubo4ka [24]

Answer:

ω₂=1.20

Explanation:

Given that

mass of the turn table ,M= 15 kg

mass of the ice ,m= 9 kg

radius ,r= 25 cm

Initial angular speed ,ω₁ = 0.75 rad/s

Initial mass moment of inertia

$I_1=\dfrac{M+m}{2}r^2$

$I_1=\dfrac{15+9}{2}\times 0.25^2\ kg.m^2$

$I_1=0.75\ kg.m^2$

Final mass moment of inertia

$I_2=\dfrac{M}{2}r^2$

$I_2=\dfrac{15}{2}\times 0.25^2\ kg.m^2$

$I_2=0.468\ kg.m^2$

Lets take final speed of the turn table after ice evaporated =ω₂ rad/s

Now by conservation angular momentum

I₁ ω₁ =ω₂ I₂

$\omega_2=\dfrac{0.75\times 0.75}{0.468}\ rad/s$

ω₂=1.20

7 0

3 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$