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zaharov [31]
3 years ago
13

25. A car is accelerating when it is (a) traveling on a straight, flat road at 50 miles per hour. (b) traveling on a straight up

hill road at 30 miles per hour. (c) going around a circular track at a steady 100 miles per hour.
Physics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

(c) going around a circular track at a steady 100 miles per hour.

Explanation:

The acceleration of an object is equal to the rate of change of velocity of the object:

a=\frac{\Delta v}{\Delta t}

where

\Delta v is the change in velocity of the object

\Delta t is the time interval

We notice that velocity is a vector, so it has both a magnitude and a direction. This means that \Delta v, the change in velocity, is either caused by a change in magnitude of velocity, or by a change in the direction.

Therefore, the acceleration is non-zero if at least one of the two is verified:

- There is a change in magnitude of the velocity

- There is a change in direction

Let's now analyze the three statements:

(a) traveling on a straight, flat road at 50 miles per hour. --> here there is no change in magnitude or direction of the velocity, so there is no acceleration.

(b) traveling on a straight uphill road at 30 miles per hour. --> here there is no change in magnitude or direction of the velocity, so there is no acceleration.

(c) going around a circular track at a steady 100 miles per hour. --> here there is no change in magnitude of the velocity, but the direction is changing (circular track), therefore there is acceleration.

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Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/s
strojnjashka [21]

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v​0 = 3.0 m/s

Displacement Δy = ?

Acceleration a = -9.8m/s2

Final velocity v = -6.0m/s

Time t=? Target unknown

We can use the kinematic formula missing Δy to solve for the target unknown t:

V=v0+at

We can rearrange the equation to solve to t:

V-v0=at

t= v-v0/a

Substituting the known value into the kinematic formula gives:

t= (-6.0m/s)-(3.0m/s)

————————————

-9.8m/s2

= -9m/s

—————-

-9.8m/s2

=0.92s

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3 years ago
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Read 2 more answers
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
Tanzania [10]

A) 18.4 m

B)

a) mass of the load

b) mass of the truck

Explanation:

A)

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g (1)

where

m is the mass of the load

\mu=0.400 is the coefficient of static friction

g=9.8 m/s^2 is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

v^2-u^2=2as (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m

B)

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck

8 0
3 years ago
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