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Natali [406]
3 years ago
9

If we want to know the velocity that an object is traveling, we must know the

Physics
2 answers:
julia-pushkina [17]3 years ago
7 0

Answer:

direction and speed.

Explanation:

Serga [27]3 years ago
4 0
If my memory serves me well, if we want to know the velocity that an object is traveling, we must know the <span>direction and speed. Velocity includes two these points listed in the previous sentence which means the answer is D.</span>
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A 2kg block has 70J of KE. It then travels 1.5 meters up a hill. As it travels up the hill friction does -12J of work on the blo
Dima020 [189]

Answer:

v = 5.34[m/s]

Explanation:

In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.

Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.

E₁ = mechanical energy at initial state [J]

E_{1}=E_{pot}+E_{kin}+E_{elas}\\

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.

In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.

E₂ = mechanical energy at final state [J]

E_{2}=E_{kin}+E_{pot}

Now we can use the first statement to get the first equation:

E_{1}+W_{1-2}=E_{2}

where:

W₁₋₂ = work from the state 1 to 2.

E_{k}=\frac{1}{2} *m*v^{2} \\

E_{pot}=m*g*h

where:

h = elevation = 1.5 [m]

g = gravity acceleration = 9.81 [m/s²]

70 - 12 = \frac{1}{2}*2*v^{2}+2*9.81*1.5

58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]

4 0
3 years ago
What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d
Butoxors [25]
Calculate the pressure due to sea water as density*depth.
That is, 
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
3 0
3 years ago
An electron of mass 9.11x10^-31 kg has an initial speed of 2.40x10^5 m/s. It travels in a straight line, and its speed increases
egoroff_w [7]

Answer: A) Force = 3.841*10^-18 N.

B) force (f) is 4.30* 10^12 times greater than the weight (Fg).

Explanation: mass of electronic charge = 9.11*10^-31kg

v = final velocity = 6.80*10^5 m/s

u = initial velocity = 2.40 * 10^5 m/s

S= distance covered = 4.8cm = 0.048m

a = acceleration

Since the acceleration of the electron is assumed to be constant, newton's laws of motion are valid.

Thus, recall that

v² = u² + 2aS

(6.80*10^5)² = ( 2.40*10^5)² + 2*a( 0.048)

46.24 * 10^10 = 5.76 * 10^10 + 0.096a

46.24 *10^10 - 5.76* 10^10 = 0.096a

40.48* 10^10 = 0.096a

a = 40.48 * 10^10/0.096

a = 4.2167*10^12m/s².

Force = mass * acceleration

Force = 9.11*10^-31 * 4.2167*10^12

Force = 3.841*10^-18 N.

Weight =Fg= mg where g = acceleration due gravity = 9.8m/s²

Fg= 9.11*10^-31 * 9.8

Fg = 8.9278* 10^-30 N

By comparing the force and the weight, we have that

F/Fg = 3.841 * 10^-18/8.9278 * 10^-30 = 4.30* 10^12.

This implies that the force (f) is 4.30* 10^12 times greater than the weight (Fg).

7 0
3 years ago
A watermelon is dropped off of a 50 ft bridge, and it explodes upon impact with the ground. How fast was it traveling in mph upo
Drupady [299]

Answer: 56.72 ft/s

Explanation:

Ok, initially we only have potential energy, that is equal to:

U =m*g*h

where g is the gravitational acceleration, m the mass and h the height.

h = 50ft and g = 32.17 ft/s^2

when the watermelon is near the ground, all the potential energy is transformed into kinetic energy, and the kinetic energy can be written as:

K = (1/2)*m*v^2

where v is the velocity.

Then we have:

K = U

m*g*h = (m/2)*v^2

we solve it for v.

v = √(2g*h) = √(2*32.17*50) ft/s = 56.72 ft/s

6 0
3 years ago
if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?
laiz [17]

Answer:

The 39.

Explanation:

8 0
2 years ago
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