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4 years ago

Can you use a machine to gain both force and speed at the same time? explain.

1 answer:

8 0

Well, It rather depends on your definition of "machine." The normal physics set of simple machines - levers, pulleys, ramps all give you increased the force at the expense of reduced speed or increased the rate at the cost of reduced force. So, no - by definition a machine is an arrangement for multiplying one while paying the cost by reducing the other. You are looking at an example of the Conservation of Energy. One of the giant rules we are pretty sure cannot be violated.<span>
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How much time does it take a power drill accelerating at 64.3 rad/s2 to achieve

Diano4ka-milaya [45]

Answer;

= 0.244 seconds

Explanation;

500 rpm is equivalent to; 1500 × 2π radians per minute

= 9424.8 rad/minute

To get revolutions per second we divide by 60

= 9424.8/60

= 157.08 radians per second.

Then we divide by 64.3 rad/s^2 to get time;

= 157.08/64.3

= 0.244 seconds.

7 0

3 years ago

Read 2 more answers

A 1,383 kg purple car is driving southbound on a road and collides with a 1,827 kg orange car, that was traveling 31.87 m/s east

user100 [1]

Answer:

Explanation:

We shall apply work energy theorem to calculate the initial velocity just after the collision .

Their kinetic energy will be equal to work done by friction .

force of friction = μ mg , where μ is coefficient of friction , m is total mass and g is acceleration due to gravity

force = .463 x 3210 x 9.8

= 14565.05 N

work done = force x displacement

= 14565.05 x 14.54 = 211775.88 J

now applying work energy theorem

1/2 m v² = 211775.88 , m is composite mass , v is velocity just after the collision

.5 x 3210 x v² = 211775.88

v² = 131.94

v 11.48 m /s

3 0

3 years ago

The tires of a car make 77 revolutions as the car reduces its speed uniformly from 95.0 km/h to 65.0 km/h. The tires have a diam

nika2105 [10]

Answer:

Explanation:

95.0 km/hr = 26.39 m/s

65 km/hr = 18.06 m/s

Circumference of a tire is 0.9π m

77 revolutions is a distance of

77(0.9π) = 69.3π m

v² = u² + 2as

a = (v² - u²) / 2s

a = (18.06² - 26.39²) / (2(69.3π))

a = -0.85 m/s²

s = (v² - u²) / 2a

s = (0² - 26.39²) / 2(-0.85)

s = 409 m

5 0

3 years ago

A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?

alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0 m$ is the final height of the ball

$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

$g=9.8 m/s^{2}$ is the acceleratio due gravity

$t$ is the time

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

$t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}$ (3)

Finally we find the time the ball is in the air:

$t=3.49 s$ (4)

7 0

4 years ago

A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an

Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

8 0

3 years ago