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lbvjy [14]
3 years ago
13

A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s^2. Find the

velocity with which the camera hits the ground.
Physics
1 answer:
cestrela7 [59]3 years ago
3 0

Answer:

42.05 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 239 m

Acceleration due to gravity (g) = 3.7 m/s²

Final velocity (v) =?

The velocity with which the camera hits the ground can be obtained as follow:

v² = u² + 2gh

v² = 0² + 2 × 3.7 × 239

v² = 0 + 1768.6

v² = 1768.6

Take the square root of both side

v = √(1768.6)

v = 42.05 m/s

Therefore, the velocity with which the camera hits the ground is 42.05 m/s

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A doubly ionized molecule (i.e., a molecule lacking two electrons) moving in a magnetic field experiences a magnetic force of 5.
Kobotan [32]

Explanation:

The given data is as follows.

           F = 5.75 \times 10^{-16} N

          q = 1.6 \times 10^{-19} C

          v = 385 m/s

       sin (63.9^{o}) = 0.876

Now, we will calculate the magnitude of magnetic field as follows.

              B = \frac{F}{qv sin (\theta)}

                  = \frac{5.75 \times 10^{-16} N}{1.6 \times 10^{-19} C \times 385 m/s \times 0.876}

                  = 0.01065 \times 10^{3} T

                  = 10.65 T

Thus, we can conclude that magnitude of the magnetic field is 10.65 T.

4 0
3 years ago
A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. Use the vertical motio
Gnoma [55]

Answer:

3.25 seconds

Explanation:

It is given that,

A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :

h=-16t^2+vt+s

Where

s is the height in feet

For the given condition, the equation becomes:

h=-16t^2+50t+7

When it hits the ground, h = 0

i.e.

-16t^2+50t+7=0

It is a quadratic equation, we find the value of t,

t = 3.25 seconds and t = -0.134 s

Neglecting negative value

Hence, for 3.25 seconds the baseball is in the air before it hits the ground.

6 0
3 years ago
Read 2 more answers
The sound of a tuba is very low. Why?
vaieri [72.5K]

Answer:

When you blow into a tuba the air vibrates very slowly.​

Explanation:

Tuba is a buzz instrument ie sound is produced in it with the help of lip vibration . It is the lowest pitched musical instrument in the brass family .

Due to absence of resonance in it , it produces music of lowest pitch , So when one blows into it the air column of the instrument vibrates very slowly producing low pitched sound.

8 0
3 years ago
A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

7 0
3 years ago
Read 2 more answers
Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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