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Margarita [4]
2 years ago
13

a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve th

is task
Physics
2 answers:
Natali [406]2 years ago
3 0
Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

Vesna [10]2 years ago
3 0

work = mgh

power = mgh/t = 63•9.8•7/5 = 864 watts

1 HP = 746 watts

so that is 834/746 = 1.2 HP

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Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 10
NNADVOKAT [17]

Answer: Their final relative velocity is -0.412 m/s.

Explanation:

According to the law of conservation,

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

Putting the given values into the above formula as follows.

      m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v

     2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v

           -4.12 \times 10^{3} kg m/s = (10^{4} kg) v

                   v = \frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}

                      = -0.412 m/s

Thus, we can conclude that their final relative velocity is -0.412 m/s.

8 0
3 years ago
The mass of a lift is 600kg.the. The maximum tensile force that the cable supporting the lift can withstand is 7kN. Calculate th
Oksi-84 [34.3K]
Force is the product of mass and acceleration .
The question is ask to find acceleration.
But acceleration is the ratio of the force and the mass.
where 600kg is the mass and 7kN is the force
NB: kilo is 1000
now we have to multiply 7N by 1000
by doing so you will have 7000N
which is the force.
Now to find the acceleration: force/ mass
which is 7000/600
therefore the maximum acceleration is 11.667


5 0
2 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

8 0
2 years ago
Assume that you can drive at a constant speed of 100 kilometers per hour. suppose you started driving from the sun. how long wou
belka [17]
The Sun is 149.6 million kilometers from the earth.
There are 8760 hours in a year. 
876000 km are traveled in a year
It would take 170.776 years to reach the sun, or 171 years rather
4 0
2 years ago
In the gene TATTCATTGTTA—TGATTT—ATTCG, CATTGTTA encodes for pepsin, a digestive enzyme. The rest of the sequence doesn’t code fo
MatroZZZ [7]

Answer:

TATTCATTCATTA—TGATTT—ATTCG

Explanation:

A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.

Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.

Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.

8 0
3 years ago
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