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arlik [135]
3 years ago
14

Please Help

Physics
1 answer:
nadezda [96]3 years ago
6 0

Answer:

The velocity and direction after 1 second is 8.1 m/s downwards

Explanation:

The equation of motion for an object in free fall can be written as follows;

v = u + g×t

Where;

v = The final velocity of the object

u = The initial velocity of the object = 0 m/s

g = The acceleration due to gravity = 9.81 m/s²

The velocity after one second is given by the velocity equation as follows;

v = 0 + 9.81 m/s² * 1 s = 9.81 m/s

The direction of the is downwards due to the attraction by the Earth's gravitational field which acts towards the Earth's center

Therefore, the velocity and direction after 1 second is 8.1 m/s downwards.

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To find the number of neutrons in an atom you would subtract?
Ghella [55]
You need to subtract the atomic number from the atomic mass to find the # of neutrons. 
8 0
3 years ago
On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per
dalvyx [7]

Answer:

a) 230 Km b) 76.7 km/h c) Please see below

Explanation:

a) If we can neglect the time while the driver accelerated, the movement can be divided in two parts, each of them at a constant speed:

x = x1 + x2  \\\\x1 = 50 km/h* 1hr = 50 km, \\x2 = 90 Km/h*2hr = 180 km\\

⇒ x = 50 km + 180 km = 230 km

b) The average x component of velocity, can be calculated applying the definition of average velocity, as follows:

vavg,x = \frac{xf-xo}{t-to}

If we choose t₀ = 0 and x₀ = 0, replacing xf and t by the values we have already found, we can find vavg,x as follows:

vavg,x =\frac{230 km}{3 hr} =76.7 km/h

c) The found value of  avg,x is not the same as the arithmetic average of the initial and final values of vx (70 Km/h) due to the time traveled at both velocities was not the same.

If the driver had droven half of the time (1.5 h) at 50 km/h and the other half at 90 km/h, total displacement would have been as follows:

x = 50 km/h*1.5 h + 90 km/h*1.5 hr = 210 km

Applying the definition of average velocity once more:

vavg,x =\frac{210 km}{3 hr} =70 km/h

which is the same as the arithmetic average of the initial and final values of vₓ.

7 0
3 years ago
What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is
ss7ja [257]

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

<u>g' = 1.09 m/s²</u>

3 0
3 years ago
Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the
Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1    ⇒   w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

           L₀ = L_f

           I₀ w₀ = I w

          w =\frac{I_o}{I} w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

          I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

        I <I₀

        I₀/I > 1    ⇒   w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0
3 years ago
How to solve a torque problem
podryga [215]
<span>you must  first select an axis of rotation about which to calculate moment arms and torques. </span>
6 0
3 years ago
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