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2 years ago

14

Why does it take double the applied force to move a mass double the size?

1 answer:

Ratling [72]2 years ago

5 0

56-999999999999999999999-4 is the best for my mom

You might be interested in

How do you measure potential and kinetic energy?

Ray Of Light [21]

Answer:

potential energy is a stored energy or energy of position (gravitational).

Kinetic energy is a energy of motion.

Explanation:

in the formula K is for the kinetic and the P stand for the potential.

3 0

3 years ago

T.E = K.E + P.E

Rasek [7]

Answer:

T.E=K.E+P.E

:146,000J=81250J+K.E

4 0

2 years ago

_____reaches the Earth’s surface through ______, then turns into ______.

VikaD [51]

Answer:

D

Explanation:

magma comes up from the core and comes out through volcanoes and then is turned into lava.

4 0

2 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

2 years ago

Ferns that eject spores generally do so in pairs, with two spores flying off in opposite directions. The structure from which th

Furkat [3]

Answer:

It cancels recoil.

Explanation:

For each action there is an equal an opposite reaction.

The principle of conservation of momentum tell us that if a single spore were ejected the fern would suffer a recoil from it. This recoil would take energy and speed from the spore. But if they are ejected in pairs the recoil is canceled and all the energy is transferred to the spores resulting in higher speeds.

5 0

3 years ago