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1 month ago

Determine the torque applied to the shaft of a car that transmits 225 hp

Physics

1 answer:

Keith_Richards [3.2K]1 month ago

8 0

Incomplete query. The complete inquiry is as follows

Calculate the torque exerted on the shaft of a vehicle transmitting 225 hp at a rotation speed of 3000 rpm.

Response:

Torque=0.51 Btu

Analysis:

Given Information

Power=225 hp

Revolutions =3000 rpm

To determine

T( torque )=?

Process

As an object is moved by force over a distance, work is performed on that object. Similarly, when torque rotates an object through an angle, work is also accomplished.

$T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }$

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Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi

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The calculation for the horizontal component is performed as follows:
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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

Keith_Richards [3271]

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, $M=75\ kg$

The mass of the ball is, $m=4\ kg$

The initial velocity of the man is, $u_m=0\ m/s(rest)$

The initial velocity of the ball is, $u_b=0\ m/s(rest)$

The final velocity of the ball is, $v_b=3.50\ m/s$

The final velocity of the man is, $v_m=?\ m/s$

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of the man and the ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Hence, the total initial momentum equals the total final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.

3 0

1 month ago

A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

kicyunya [3294]

Answer:

$\frac{V_{big}}{V_{small}} = n^{2/3}$

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

$V_{big} = \frac{KQ}{R}$

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

$\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)$

$r = \frac{R}{n^{1/3}}$

Now, the potential for the smaller droplets is given as:

$V_{small} = \frac{kq}{r}$

$V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}$

$V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}$

$\frac{V_{big}}{V_{small}} = n^{2/3}$

7 0

2 months ago

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential o

Maru [3345]

Answer:

The excess charge is $Q = 3.5 *10^{-7} \ C$

Explanation:

According to the question, we are informed that

The diameter is $d = 45 \ cm = 0.45 \ m$

The potential of the surface is $V = 14 \ kV = 14 *10^{3} \ V$

The radius of the sphere is

$r = \frac{d}{2}$

by plugging in given values

$r = \frac{0.45}{2}$

$r = 0.225 \ m$

The potential at the surface is mathematically expressed as

$V = \frac{k * Q }{r }$

Where k is Coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

Based on the question stating there are no other charges, Q represents the excess charge

Therefore

$Q = \frac{V* r}{ k}$

inserting the numerical values

$Q = \frac{14 *10^{3} 0.225}{ 9*10^9}$

$Q = 3.5 *10^{-7} \ C$

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29 days ago