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Elza [17]
2 years ago
8

A 65 kg woman is horizontal in a push-up position. What are the vertical forces acting on her hands?​

Physics
1 answer:
Leni [432]2 years ago
8 0

Answer:

Explanation:

Sum moments about her toes, to zero.

65(9.8)[1.00] - F[1.50] = 0

F = 424.666... N

Rounding to two significant digits makes it

F = 420 N

Don't know what to tell you, None of those options seem really close and you have two options at 410.4 N so I'm guessing a typo.

If you only take a single significant digit on gravity to make it 10 m/s² you end up with answer B), but why would one assume four significant digits in an answer when only a single significant digit went into the arguments? Only thing we can say for certain is it's probably not C)

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Explain what is the meaning of positive connotation and negative connotation.
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In a two-body collision, if the momentum of the system is conserved, then which of the
lakkis [162]

Answer:

c) may also be conserved

Explanation:

Momentum is conserved  in both elastic and inelastic type of collisions.

But the differences is that:

In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.

whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.

So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.

Hence, may also be conserved is the appropriate option here.

5 0
3 years ago
A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6
andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using A\int_0^{12}\left(B+Cx)dx

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6,

hence, the mass of the rod is 126.6 g.

7 0
3 years ago
A 21 g ball is shot from a spring gun whose spring has a force constant of 579 N/m. The spring can be compressed 1 cm. How high
vodka [1.7K]

Answer:

The answer  to the question is

The ball will go 0.14 meters high if the gun is aimed vertically

Explanation:

The energy in the spring → Energy, E = \frac{1}{2}kx^2

Where E = energy in the spring

k = Spring constant

x = Spring compression or stretch

Therefore E = \frac{1}{2} 579*0.01^2 = 0.02895 J

The spring energy is transferred  to the ball as kinetic energy based on the first law of thermodynamics which states that energy is neither created nor destroyed

Kinetic energy = KE = \frac{1}{2}mv^2

From which v = \sqrt{\frac{2*KE}{m} } = \sqrt{\frac{2*0.02895}{0.021} } = 1.66 m/s

from v² =u² - 2·a·S

Where v = final velocity = 0 m/s

u = initial velocity = 1.66 m/s

a = g = Acceleration due to gravity

S = height

Therefore 0 = 1.66² - 2×9.81×S

or S = 1.66² ÷ (2×9.81) = 0.14 m

7 0
3 years ago
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