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kumpel [21]
2 years ago
6

A 28-kg particle exerts a gravitational force of 8.3 x 10^-9 N on a particle of mass m, which is 3.2 m away. What is m? A) 140 k

g B) 8.5 x 10^-10 kg C) None of the choices are correct D) 46 kg E)1300 kg
Physics
1 answer:
xxTIMURxx [149]2 years ago
6 0

Answer:

Mass of another particle, m = 46 kg  

Explanation:

it is given that,

Mass of first particle, m₁ = 28 kg

Gravitational force, F=8.3\times 10^{-9}\ N

Distance between the particles, d = 3.2 m

We need to find the mass m of another particle. It is given by the formula as follows :

F=G\dfrac{m_1m}{d^2}

m=\dfrac{Fd^2}{Gm_1}

m=\dfrac{8.3\times 10^{-9}\ N\times (3.2\ m)^2}{6.67\times 10^{-11}\times 28\ kg}

m = 45.5 kg

or

m = 46 kg

So, the correct option is (d) "46 kg". Hence, this is the required solution.

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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
1 year ago
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of
adoni [48]

Answer:

3311N

Explanation:

r = radius = 600m

V = speed = 150m/s

Mass = weight = 70kg

The weight of pilot when calculated due to circular motion

W = tv

Fv = mv²/r

Fv = 70x150²/600

Fv = 79x22500/600

= 15750000/600

= 2625N

Real Weight of the pilot = m x g

= 70 x 9.8

= 686N

The apparent Weight is calculated by

Mv²/r + mg

= 2625N + 686N

= 3311 N

Therefore the apparent Weight is 3311N

6 0
1 year ago
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
The deuterium nucleus starts out with a kinetic energy of 1.24 × 10-13 joules, and the proton starts out with a kinetic energy o
MrMuchimi

Answer:

The total kinetic energy of both particles is 2.43\times10^{-13}

Explanation:

Given that,

Kinetic energy of nucleusK.E= 1.24\times10^{-13}\ J

Kinetic energy of proton K.E= 2.47\times10^{-13}\ J

Radius of proton r= 0.9\times10^{-15}\ m

We need to calculate the final potential energy

Using formula of final potential energy

U=\dfrac{kq^2}{r}

Put the value into the formula

U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{2\times0.9\times10^{-15}}

U_{f}=1.28\times10^{-13}\ J

We need to calculate the initial energy of both the particles

Using formula of energy

E_{i}=(K.E_{n}+K.E_{p})+U_{i}

E_{i}=1.24\times10^{-13}+2.47\times10^{-13}+0

E_{i}=3.71\times10^{-13}\ J

We need to calculate the total kinetic energy of both particles

Using conservation of energy

E_{i}=E_{f}

E_{i}=K.E_{f}+U_{f}

3.71\times10^{-13}=K.E_{f}+1.28\times10^{-13}

K.E_{f}=3.71\times10^{-13}-1.28\times10^{-13}

K.E_{f}=2.43\times10^{-13}

Hence, The total kinetic energy of both particles is 2.43\times10^{-13}

3 0
2 years ago
Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder
Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0
2 years ago
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