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Lubov Fominskaja [6]
3 years ago
8

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

Physics
1 answer:
crimeas [40]3 years ago
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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Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of
morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner(v_1)=3.1 m/s

speed of second runner(v_2)=4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner t=\frac{295.6}{4.65}=63.56 s

time required by first runner to reach finish line=\frac{250}{3.1}=80.64 s

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0
3 years ago
A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?
kramer

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

3 0
3 years ago
Read 2 more answers
According to recent typical test data, a Ford Focus travels 0.240 mi in 19.3 s, starting from rest. The same car, when braking f
Anit [1.1K]

Answer:

Explanation:

a )

While breaking initial velocity u = 62.5 mph

= 62.5 x 1760 x 3 / (60 x 60 )  ft /s

= 91.66 ft / s

distance trvelled s = 150 ft

v² = u² - 2as

0 = 91.66²  - 2 a x 150

a = - 28 ft / s²

b ) While accelerating initial velocity u = 0

distance travelled s = .24 mi

time = 19.3 s

s = ut + 1/2 at²

s is distance travelled in time t with acceleration a ,

.24 = 0 + 1/2 a x 19.3²

a = .001288 mi/s²

= 2.06 m /s²

c )

If distance travelled s = .25 mi

final velocity v = ? a = .001288 mi / s²

v² = u² + 2as

= 0 + 2 x .001288 x .25

= .000644

v = .025 mi / s

= .0025 x 60 x 60 mi / h

= 91.35 mph .

d ) initial velocity u = 59 mph

= 86.53 ft / s

final velocity = 0

acceleration = - 28 ft /s²

v = u - at

0 = 86.53 - 28 t

t = 3 sec approx .

4 0
3 years ago
As a pendulum swings from its highest to lowest position, what happens to its kinetic and potential energy?
DIA [1.3K]

The potential energy decreases while the kinetic energy increases.



7 0
3 years ago
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The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet
kirill [66]
The answer would be 54 m/s as the maximum speed
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