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sergejj [24]
3 years ago
10

Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that l

oad in 1 6 the time it takes the other. How much more power does the faster crane have?
Physics
1 answer:
Neporo4naja [7]3 years ago
5 0

Answer:

Explanation:

Given

Load W=20000\ N

height to which load is raised h=250\ m

Another crane take \frac{1}{6} th time to lift the load

Energy required required to lift the Weight

E=W\times h

E=20000\times 250

E=5,000,000\ J

Suppose P_1 and P_2 is the Power required to lift the weight in t and \frac{t}{6} time

E=P_1\times t

E=P_2\times \frac{t}{6}

P_1\times t=P_2\times \frac{t}{6}

thus

P_2=6P_1

Second Crane requires 6 times  more power than the slow crane                                              

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Suppose you have a 115 kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood sur
alexdok [17]

Answer:

maximum force (in N) can you exert horizontally on the crate without moving it = 564.075 N

Explanation:

Given data:

mass m= 115 kg

coefficient of friction μ =0.5

according to newton law of motion the net force acted on body is given by

F_{net}=ma=F_{max}- F_{f}

here the body is in rest ,

F_{net}=0\\\Rightarrow F_{max}=F_{f}\\\text{frriction force}  F_{f} = \mu\times N\\

N= it is the normal reaction force

N= mg = 115×9.81

therefore,

F_{f} = 0.5\times115\times9.81 = 564.075 N = F_{max}

4 0
3 years ago
An incident light ray strikes water at an angle of 20 degrees. The index of refraction of air is 1.0003, and the index of refrac
sammy [17]
In order to determine the angle of the refracted ray, we may apply Snell's law, which states that the ratio of the sines of the angles of incidence and refraction is constant for a given wave when it passes through two different media. Mathematically, this is:
n₁sin(∅₁) = n₂sin(∅₂)
Where n is the refractive index. Substituting the values given into the equation:
1.0003 * sin(20°) = 1.33 * sin(∅)
∅ = 14.91

The angle of the refracted ray is 15°.
4 0
3 years ago
Read 2 more answers
The amount of WORK done is determined by 2 factors.
RUDIKE [14]

Explanation :

Work is done when a force is applied to create a displacement on an object.

Thus, the work done depends on the two factors i.e.

(1) Applied force (F)

(2) Distance or displacement (d)

Mathematically, work done is W= F.s

It also depends on the angle between the force and the displacement.

        W=Fs\ cos\ \theta

For example,

A person carries a weight of 20 kg and lifts it on his head 1.5 m above the surface. So, the work done by him on the luggage will be:

W= F\times s

or

W=m\times g \times s

W=20\ kg\times 9.8\ m/s^2\times 1.5\ m

So, W=294\ Joules

Hence, the work done by him on the luggage is 294 Joules.

6 0
3 years ago
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
A 52.0-kg person, running horizontally with a velocity of +3.63 m/s, jumps onto a 15.2-kg sled that is initially at rest. (a) Ig
trasher [3.6K]

Answer:

The coefficient of kinetic friction between the sled and the snow is 0.0134

Explanation:

Given that:

M = mass of person = 52 kg

m = mass of sled = 15.2 kg

U = initial velocity of person = 3.63 m/s

u = initial velocity of sled = 0 m/s

After collision, the person and the sled would move with the same velocity V.

a) According to law of momentum conservation:

Total momentum before collision = Total momentum after collision

MU + mu = (M + m)V

V=\frac{MU+mu}{M+m}

Substituting values:

V=\frac{MU+mu}{M+m}=\frac{52(3.63)+15.2(0)}{52+15.2} =2.81m/s

The velocity of the sled and person as they move away is 2.81 m/s

b) acceleration due to gravity (g) = 9.8 m/s²

d = 30 m

Using the formula:

V^2=2\mu(gd)\\\mu=\frac{V^2}{2gd} \\\mu=\frac{2.81^2}{2*9.8*30} =0.0134

The coefficient of kinetic friction between the sled and the snow is 0.0134

3 0
3 years ago
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