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Bingel [31]
3 years ago
6

1. A 61-kg woman doing pull-ups lifts her body a distance of 0.32 meters in 1.8 seconds. What is the power provided by her bicep

s?
Physics
1 answer:
Allushta [10]3 years ago
4 0

Answer:

P = mgh/t = 61(9.8)(0.32)/1.8 = 106.275555... ≈ 110 W

Explanation:

Power is the rate of doing work.

The work changes her potential energy.

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What is the velocity of a wave with a wavelength of 9 meters and a period of 0.006
Ipatiy [6.2K]

Answer: 1,500m/s

Explanation:

Relationship existing between velocity of a wave (v), wavelength(¶) and frequency(f) is

v = f¶... (1)

Since Frequency (f) is the reciprocal of the period (T);

Frequency = 1/Period i.e F = 1/T... (2)

Substituting equation 2 into 1 we have;

v = 1/T × ¶

v = ¶/T

Given wavelength ¶ = 9m

Period T = 0.006s

v = 9/0.006

v = 1,500m/s

The velocity of the wave will be 1,500m/s

8 0
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99% of all the matter that can be observed in the universe exists as
Temka [501]
Has mass and occupied space
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4 years ago
An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa
Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

F=G*\frac{m_a*m_p}{r^2}

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2

so:

m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg

7 0
4 years ago
What is the equation of instantaneous velocity<br>​
Oksi-84 [34.3K]

Answer:

it's possible to calculate an object's velocity at any moment along its path. This is called instantaneous velocity and it is defined by the equation v = (ds)/(dt),in other words, the derivative of the object's average velocity equation.

7 0
3 years ago
In April 1974, Steve Prefontaine completed a 10 km race in a time of 27 min, 43.6 s. Suppose "Pre" was at the 8.13 km mark at a
SOVA2 [1]

Answer:0.084 m/s^2

Explanation:

Given

Total time=27 min 43.6 s=1663.6 s

total distance=10 km

Initial distance d_1=8.13 km

time taken=25 min =1500 s

initial speed v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s

after 8.13 km mark steve started to accelerate

speed after 60 s

v_2=v_1+at

v_2=5.6+a\times 60

distance traveled in 60 sec

d_2=v_1\times 60+\frac{a60^2}{2}

d_2=336+1800 a

time taken in last part of journey

t_3=1663.6-1560=103.6 s

distance traveled in this time

d_3=v_2\times t_3

d_3=\left ( 5.6+a\times 60\right )103.6

and total distance=d_1+d_2+d_3

10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6

1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6

a=0.084 m/s^2

5 0
3 years ago
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