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2 years ago

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Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat

generated by the machine, and all other types of work done by machine, ( blank) J of output work will be done.

2 answers:

Softa [21]2 years ago

8 0

Answer:

Cori uses 475 J of energy from her muscles to push a bar 1 m on a weight machine at the gym. Between the bar’s motion , the heat generated by the machine, and all other types of work done by machine, ( 475 ) J of output work will be done.

Explanation:

on edge

dimulka [17.4K]2 years ago

5 0

Answer:

475

Explanation:

Cori does not exert any more force than 475 J, so 475 is the answer.

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A squirrel in a tree drops an acorn. how long does it take the acorn to fall 20 feet?

mart [117]

We use the equation of motion,

$S= ut+\frac{1}{2}at^{2}$

Here, S is the height, u is initial velocity and a is acceleration.

Given, $S = 20 \ ft$ $S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft} = 6.096 \ m$

As acorn falls from tree, therefore we take the value of $a = 9.8 \ m/s^2$ and initial velocity $u = 0$ .

Substituting these values in equation of motion,

$6.096 \ m = 0 \times t +\frac{1}{2} \times 9.8 m/s^2 (t)^2 \\\\\ t = 1.12 \ s$

Thus, the time taken by the acorn to fall 20 feet ( 6.096 m ) is 1.12 s.

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2 years ago

A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha

Diano4ka-milaya [45]

To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

$p = m\Delta v$

Here,

m = mass

$\Delta v =$ Change in velocity

As we can see there are two types of velocity at the moment the object makes the impact,

the first would be the initial velocity perpendicular to the wall and the final velocity perpendicular to the wall.

That is to say,

$v_i = vcos\theta$

$v_f = -v sin\theta$

El angulo dado es de 45° y la velocidad de 25, por tanto

$v_i = (25)cos(45) = 17.68m/s$

$v_f = -(25)sin(45) = -17.68m/s$

The change of sign indicates a change in the direction of the object.

Therefore the impulse would be as

$p = 0.060(-17.68-17.68)$

$p = -2.12kg \cdot m/s$

The negative sign indicates that the pulse is in the opposite direction of the initial velocity.

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2 years ago

Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B

gregori [183]

Answer:

The force P required is 1759.22 N

Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

To start with determining the value of the angle ∝;

$tan \ \alpha = \dfrac{CD}{BD}$

where;

BD = AB-AD

Then;

$tan \ \alpha = \dfrac{CD}{AB-AD}$

$\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )$

replacing their respective values, where;

CD = 2 sin 30° m, AB = 2m and AD = 2 cos 30° m

$\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{2-1.732} \bigg )$

$\alpha = tan^{-1} \bigg(\dfrac{1}{0.268} \bigg )$

$\alpha = tan^{-1} \bigg(3.73\bigg )$

$\alpha \simeq 75^0$

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

$T_{BC} = P$

Using the force equilibrium expression along the horizontal direction.

$\sum F_x = 0\\\\ -T_{AC} \ cos \ 30^0 + Pcos \alpha = 0$

replacing the value of $\alpha \simeq 75^0$

$-T_{AC} \ cos 30^0 + P cos 75^0 = 0$

$P \ cos \ 75^0 = T_{AC} \ cos \ 30^0$

$P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - - (1)$

Along the vertical direction, the force equilibrium equation can be expressed as:

$\sum F_y =0$

$-W + P \ sin \alpha + T_{AC} \ sin \ 30^0 = 0$

$W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0$

replacing $\alpha \simeq 75^0$ and $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

Also, replacing W for (200 × 9.81) N

$200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0$

$200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0$

$1962= T_{AC} \ ( cos \ 30^0 \ tan \ 75^0 + \ sin \ 30^0)$

$1962= T_{AC} \ (0.8660\times 3.732 + 0.5)$

$1962= T_{AC} \ (3.231912 + 0.5)$

$1962= T_{AC} \ (3.731912)$

$T_{AC} = \dfrac{1962}{ \ (3.731912)}$

$T_{AC} = 525.736 \ N$

From $P =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \ cos \ 30^0}{\ cos \ 75^0}$

$P =\dfrac{ 525.736 \times0.866}{0.2588}$

P = 1759.22 N

Thus, the force P required is 1759.22 N

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1 year ago

A clothes dryer in a home has a power of 4,500 watts and runs on a special 220-volt household circuit,

ladessa [460]

1). <u>Power = (voltage)² / (Resistance)</u>

   4,500 = (220)² / Resistance

Multiply each side by (resistance) : 4,500 x resistance = (220)²

Divide each side by 4,500 : Resistance = (220)² / 4,500 = <em>10.76 ohms</em>

2). <u>Power = (voltage) x (Current)</u>

Divide each side by (voltage): Power / voltage = Current

      4,500 / 220 = <em>20.45 Amperes</em>

3). 4,500 watts = 4.5 kilowatts

   (4.5 kilowatts) x (4 hours) = <em>18 kilowatt-hours</em>

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2 years ago

A segment of wire of total length 2.0 m is formed into a circular loop having 5.0 turns. If the wire carries a 1.2-A current, de

docker41 [41]

Answer:

Magnetic field at the center of the loop $B=5.89\times 10^{-5}\ T.$

Explanation:

It is given that total length of wire is 2 m and number of circular loop is 5 turns.

Therefore ,

$5\times ( 2\pi r)=2 \ m .\\\\r=\dfrac{1}{5 \pi}=0.064\ m.$

We know , magnetic field at the center of loop is given by :

$B=N\dfrac{\mu_o i}{2r}$

Putting all values in above equation we get :

$B=5\times \dfrac{4\pi\times 10^{-7}\times 1.2}{2\times 0.064}\\\\B=5.89\times 10^{-5}\ T.$

Hence , this is the required solution.

8 0

2 years ago