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kirza4 [7]
3 years ago
7

Students in a physical education class practiced rope climbing. Students gripped a rope that hung from the ceiling and lifted th

emselves up the rope using only their arms. What must be measured to determine the power output of the climbers?
A) tension on the rope and the final height
B) body mass, final height, and the tension on the rope
C) tension on the rope and the time required to complete the climb
D) body mass, final height, and the time required to complete the climb
Physics
2 answers:
Lelechka [254]3 years ago
6 0
Body mass final height and the tension on the rope
ahrayia [7]3 years ago
5 0

Answer:

D) body mass, final height, and the time required to complete the climb

Explanation:

Power is defined as the rate at which a work is done

So there power out put could be measured using the mass of the climber hieght reached with time

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Convert 68852 millijoules into Calories. (Write your answer in the decimal form. Do not include units in your answer).
alex41 [277]

Answer: 68852 millijoules = 16.46 calories

Explanation:

Given;

Convert 68852 millijoules to calories.

1 calorie = 4.184J = 4184millijoules

Therefore,

1 millijoule = 1/4184 calories

68852 millijoule = 68852 × 1/4184 calories

= 16.46 calories

6 0
3 years ago
If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. Wha
emmainna [20.7K]

Answer:

B

Explanation:

4 0
3 years ago
(1 point) A projectile is fired from ground level with an initial speed of 600 m/sec and an angle of elevation of 30 degrees. Us
konstantin123 [22]

Answer:

(a) The range of the projectile is 31,813.18 m

(b) The maximum height of the projectile is 4,591.84 m

(c) The speed with which the projectile hits the ground is 670.82 m/s.

Explanation:

Given;

initial speed of the projectile, u = 600 m/s

angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m

(b) The maximum height of the projectile in meters;

H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m

(c) The speed with which the projectile hits the ground is;

v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s

5 0
3 years ago
A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire
oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0
3 years ago
A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du
Lana71 [14]

Answer:

a_E=\dfrac{Gm}{r^2}

Explanation:

M_E = Mass of the Earth =  5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}

The acceleration that the Earth will feel is a_E=\dfrac{Gm}{r^2}

7 0
3 years ago
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