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Paraphin [41]
3 years ago
15

A pulley system lifts a 500-lb. block 2.0 feet with an effort of only 50 lb. If the 50 lb. moves 30 feet, calculate the efficien

cy of the pulley system.
Physics
2 answers:
dybincka [34]3 years ago
6 0

Answer:

Efficiency of the pulley system is 66.7 %

Explanation:

It is given that,

Input force, F_{in}=500\ lb

Distance, d₁ = 2 feet

Output force, F_{out}=50\ lb    

Distance, d₂ = 30 feet

The efficiency of pulley system is defined as ratio of output work done to the input work done i.e.

\eta=\dfrac{W_{out}}{W_{in}}

\eta=\dfrac{500\times 2}{50\times 30}\times 100

\eta=66.7\%      

Hence, the efficiency of the pulley system is 66.7 %    

tatiyna3 years ago
4 0
The answer to this is 67%
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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve
Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision  = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2

v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96

Therefore, the velocity of tennis racket after collision is 14.96m/s

7 0
3 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
HELP!!
tresset_1 [31]

Answer:

So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3

3 0
2 years ago
of 72.0 units. If the mass of Object 1 is halved AND the mass of object 2 is tripled, then the new gravitational force will be u
katen-ka-za [31]

Answer:

youtg jugre contact coming call details

6 0
2 years ago
A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 146 cm and makes an angl
svetlana [45]

Answer:

B = 191.26 cm

θ = -14.73°

Explanation:

given,

magnitude of the first displacement(A) = 146 cm

at an angle of 124°

resultant magnitude = 137 cm

and angle made with x-axis by the resultant(R) = 32.0°

component of A in X and Y direction

A x = A cos θ  = 146 cos 120° = -73 cm

A y = A sin θ = 146 sin 120° = 126.4 cm

now component of resultant in x and y direction

R x = 137 cos 35°

    = 112.2 cm

R y = 137 sin 35°

     = 78.6 cm

resultant is the sum of two vectors

R = A + B

R x = A x + B x

B x =  112.2 - (-73) = 185.2 cm

B y = R y - A y

B y = 78.6 - 126.4 = -47.8 cm

magnitude of B

B = \sqrt{B_x^2+B_y^2}

B = \sqrt{185.2^2+-47.8^2}

B = 191.26 cm

angle\theta = tan^{-1}\dfrac{-47.8}{185.2}

θ = -14.73°

6 0
3 years ago
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