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Maslowich
3 years ago
9

Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth rad

ii from earth's center).
Physics
1 answer:
Anettt [7]3 years ago
5 0

Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, d=1.9\times 10^7\ m

Mass of Earth, M=5.97\times 10^{24}\ kg

Radius of Earth, r=6.37\times 10^6\ m

We need to find the force of gravity above the surface of Earth. It is given by :

F=G\dfrac{mM}{R^2}

R = r + d

R = 25370000 m

F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

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Do you think extremist groups, cults, and gangs pose a significant threat in your country? Why or why not?
Lynna [10]

Answer:

let me check the answer for you

4 0
2 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
24- Which of the fallowing devices make use a lens
mezya [45]

Answer:

d.

Explanation:

all because the projector is a lens and the magnifying also the microscope

6 0
3 years ago
Read 2 more answers
A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the
REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0
3 years ago
A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
svetoff [14.1K]

Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

8 0
3 years ago
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