I need help on this I'm not the best at sicence

A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10 m/s. The instan

gizmo_the_mogwai [7]

Answer:

The velocity just before hitting the ground is $v_f = 30 m/s$

Explanation:

From the question we are told that

The initial speed is $u = 10 m/s$

The final speed is $v = 30 \ m/s$

From the equations of motion we have that

$v^2 =u^2 + 2as$

Where s is the distance travelled which is the height of the cliff

So making it the subject of the the formula we have that

$s = \frac{v^2 - u^2 }{2a}$

Where a is the acceleration due to gravity with a value $a = 9.8m/s^2$

So

$s = \frac{30^2 - 10^2 }{2 * 9.8 }$

$s = 40.8 \ m$

Now we are told that was through horizontally with a speed of

$v_x =10 m/s$

Which implies that this would be its velocity horizontally through out the motion

Now it final velocity vertically can be mathematically evaluated as

$v_y = \sqrt{2as}$

Substituting values

$v_y = \sqrt{(2 * 9.8 * 40.8)}$

$v_y = 28.3 \ m/s$

The resultant final velocity is mathematically evaluated as

$v_f = \sqrt{v_x^2 + v_y^2}$

Substituting values

$v_f = \sqrt{10^2 + 28.3^2}$

$v_f = 30 m/s$

5 0

3 years ago

An indirect measurement of the speed of molecules is?

Arlecino [84]

Answer:

An indirect measurement of the speed of molecules is temperature

3 0

4 years ago

Read 2 more answers

3. How did Kepler's work influence Newton?

Katarina [22]

Johannes Kepler and his laws were a great influence on Isaac Newton. ... Newton used his laws of gravity and motion to derive Kepler's laws and show that the motion of the planets could be explained using mathematics and physics.

3 0

3 years ago

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suppose that the man pictured on the front side is orbiting the earth (mass=5.98 x 10^24kg at a distance of 310 miles above the

wlad13 [49]

Answer:

a = 9.81[m/s^2]; v = 18683.5[m/s]

Explanation:

The stament of the problem is:

Suppose that the man pictured on the front side is orbiting the earth (mass = 5.98 x 1024kg) at a distance of 310 miles (1600 meters = 1 mile) above the surface of the earth (radius = 4000 miles).

a. What acceleration does he experience due to the earth's pull?

b. What tangential velocity must he possess in order that he orbit safely (in m/s)?

First we need to convert all the initial data to units of the SI

Rs = 310 [mil] = 498897 [m]

RT= 400 [mil] = 643738 [m]

r = Rs + RT = 1142635 [m] "distance from the center of the earth to the man"

$F=G*\frac{M*m}{r^{2} } \\where:\\$

G = universal gravitational constant

M = mass of the earth [kg]

m = mass of the man [kg]

r = distance from the center of the earth to the man [m]

a)

The acceleration he is experimenting is the same acceleration given by the gravity, therefore:

a = g = 9.81[m/s^2]

b)

To find the tangential velocity, we must determinate the force exerted by the earth.

Now we will find the force exerted by the gravity when the man is orbiting the earth at distance r.

$G = 6.67*10^{-11} [\frac{N*m^{2} }{kg^{2} } ]\\M=5.98*10^{24}[kg]\\ m=100 [kg]\\Replacing:\\F = G*\frac{M*m}{r^{2} }$

$F = 6.67*10^{-11}*\frac{5.98*10^{24}*100 }{1142635^{2} } \\ F= 30550 [N]$

And this force will be equal to the following expression:

$F = m*\frac{v^{2} }{r} \\where:\\v= tangential velocity [m/s]\\v=\sqrt{\frac{F*r}{m} } \\v=\sqrt{\frac{30550*1142635}{100} } \\v=18683.5[m/s]$

8 0

3 years ago

(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less

Free_Kalibri [48]

Answer:

q = 7.4 10⁻¹⁰ C

Explanation:

a) The magnetic force is given by the expression

F = q v x B

Where the blacks indicate vectors, q is the electric charge, v at particle velocity and B the magnitude of the magnetic field. If the velocity is perpendicular to the magnetic field, the sine is 1

F = q v B

Let's calculate the charge

q = F / vB

q = 1.00 10⁻¹² / 30.0 B

For the magnetic field of the earth we have a value between 25μT and 65μT, an intermediate value would be 45 μT, let's use this value.

q = 1 10⁻¹² / (30 45 10⁻⁶)

q = 7.4 10⁻¹⁰ C

b) In laboratories and modern electronics, currents of up to 1 10⁻⁶ A can be achieved without much difficulty, in advanced and research laboratories currents of up to 1 10⁻¹² can be managed. Load values (coulomb) cannot they are widely used today for work, but 1 mA = 3.6C, so we see that getting loads with the value of 10⁻¹⁰ C implies very small current less than 1 10⁻¹³ A, which only in laboratories of Very specialized can be created. Consequently, from the above it would be difficult to find loads lower than the calculated

The electrostatic charge is the one created by the friction between two surfaces, it is an indicated charge, in this case it would be possible to have better wing loads found from 10⁻¹⁰C

4 0

3 years ago