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3 years ago

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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?

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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A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

olga_2 [115]

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, $a=3.1\ m/s^2$

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

$v=u+at$

$v=0+3.1\times 7.9$

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

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3 years ago

Does friction always oppose relative motion?

Ilya [14]

Answer:

Yes

Explanation:

Friction is a force that opposes relative motion between systems in contact. One of the simpler characteristics of friction is that it is parallel to the contact surface between systems and always in a direction that opposes motion or attempted motion of the systems relative to each other.

3 0

3 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

3 years ago

The joule and the kilowatt-hour are both units of energy. 15 kw · h is equivalent to how many joules? answer in units of j.

choli [55]

The solution for the problem is:

1 Watt = 1 Joule per second
1 Watt*second = 1 Joule

a Kilowatt is 1,000 Watts
an hour is 60 seconds times 60 minutes or 3,600 seconds
a Kilowatt * hour is 1,000 Watts in 3,600 seconds

15 W*h = 15,000 Watt*hour = 15,000 Watt * 3,600 seconds = 54,000,000 Watt*second

54,000,000 Watt*second = ? Joules
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3 0

3 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago