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3 years ago

Is the distance on a round-trip positive, negative, or zero?

Physics

2 answers:

ELEN [110]3 years ago

5 0

It should be positive

mars1129 [50]3 years ago

4 0

On a round trip, where you return to the place you started from, the distance for the whole trip is positive, and the displacement is zero.

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Consider a uniform hoop of radius r and mass m rolling without slipping. which is larger, its translational kinetic energy or it

OlgaM077 [116]

translational kinetic energy is larger than its rotational kinetic energy

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3 years ago

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.

Read more: brainly.com/question/8898885

8 0

2 years ago

Which statement is true according to Newton's first law of motion?

katrin2010 [14]

C) In the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

hope this helps and have a great day :)

8 0

3 years ago

Read 2 more answers

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

Leokris [45]

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

$T=m\frac{v^2}{r}$

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

$m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg$

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

$T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N$

5 0

3 years ago

Rearrange the formula for mechanical energy to solve for height:

docker41 [41]

Explanation:

Given formula:

ME= $\frac{1}{2}$ mv²+mgh

To make height the subject of the formula, follow the following procedures;

Subtract $\frac{1}{2}$ mv² from both side of equation

M.E - $\frac{1}{2}$ mv² = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mv²+mgh

This gives:

M.E - $\frac{1}{2}$ mv² = mgh

Multiply both sides of the expression by $\frac{1}{mg}$

( M.E - $\frac{1}{2}$ mv² ) x $\frac{1}{mg}$ = $\frac{1}{mg}$ x mgh

h = ( M.E - $\frac{1}{2}$ mv² ) x $\frac{1}{mg}$

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

4 0

4 years ago