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yaroslaw [1]
3 years ago
8

A weightlifter is attempting a biceps curl with a 200 n barbell. the moment arm of the barbell about the elbow joint is 40 cm. t

he moment arm of the elbow flexor muscles is 4 cm. how much force must be produced by these muscles to hold the 200 n barbell in static equilibrium?
Physics
1 answer:
CaHeK987 [17]3 years ago
4 0
Weight of the barbell W = 200 Ndistance of the joint is r = 40 cm = 0.4 mtorque created by the weight at the joint is                  τ = F*r                     = 200 N*0.4 m                     = 80 N.mat equilibrium condition ,    Στ = force*distance - 80 N.m = 0             F'*0.4 - 80 N.m = 0             F'*0.4 = 80          force F' = 200 N
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netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

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6 0
3 years ago
A 12 V automobile battery is connected to an electric starter motor. The current
krek1111 [17]

Answer: 3000\ W,\ 78,000\ J

Explanation:

Given

The voltage of the battery is V=12\ V

Current through the battery is I=250\ A

(a) Power is the product of voltage and current

Power dissipates is given by

P=VI\\P=12\times 250\\P=3000\ W

(b) Energy delivered

E=P\times t

E=3000\times 26\\E=78,000\ J

3 0
3 years ago
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