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2 years ago

I need help with nine and 10 I will really appreciate it

Physics

1 answer:

horsena [70]2 years ago

7 0

Answer:

9] V = D ÷ T

Take any distance value from the graph and its relevant time.

V = 4 ÷ 2

V = 2 m/s

[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]

10] Take the distance value and its time for the highest peak of B.

V = 20 ÷ 2

V = 10 m/s

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A 5.31 kg object is swung in a vertical circular path on a string 2.99 m long. The acceleration of gravity is 9.8 m/s 2 . If the

11111nata11111 [884]

Answer:

T = 120.3 N

Explanation:

Since, the tension in the rope is acting against both the centripetal force and the weight of the stone. As both act downward towards center of the circle and tension acts towards point of support that is upward. So, tension will be equal to the sum of centripetal force and weight of the stone:

Tension = Centripetal Force + Weight of Stone

T = mv²/r + mg

where,

m = mass of stone = 5.31 kg

r = radius of circle = length of string = 2.99 m

g = 9.8 m/s²

Therefore,

T = (5.31 kg)(6.2 m/s)²/(2.99 m) + (5.31 kg)(9.8 m/s²)

T = 68.27 N + 52.03 N

T = 120.3 N

4 0

3 years ago

PLEASE HELP HDJFJDJDJJCJDJDJFJD

GenaCL600 [577]

Answer:

Explanation:

free

5 0

3 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

A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta

Alex

Answer:

The depth will be equal to 6141.96 m

Explanation:

pressure on the submarine $P_{sea}$ = 62 MPa = 62 x 10^6 Pa

we also know that $P_{sea}$ = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

$P_{sea}$ = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine $P_{g}$ = 101 kPa = 101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure $P_{sea}$ , we have

62 x 10^6 = 10094.49h

depth h = 6141.96 m

7 0

3 years ago

Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0

3 years ago