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3 years ago

2)

1 answer:

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B). A <span>car that rounds a curve at a constant speed is accelerating.

</span><span>D). A car that is set to a constant speed of 60 miles per hour is
accelerating IF the road ever curves. </span><span>

</span>

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(a) Calculate the height of a cliff if it takes 2. 35 s for a rock to hit the ground when it is thrown straight up from the clif

Natali [406]

(a) The height of the cliff will be 8.26 meters.

(b) The time would it take to reach the ground will be 0.717 sec.

<h3>What is velocity?</h3><h3 />

The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component.

(a) The height of the cliff will be 8.26 meters.

According to Newton's second equation of motion

$\rm H =ut-\frac{1}{2} gt^2 \\\\ \rm H =8\times 2.35-\frac{1}{2} 9.81 (2.35)^2\\\\\rm H =8.16 \; m$

Hence The height of the cliff will be 8.26 meters.

(b)The time would it take to reach the ground will be 0.717 sec.

We must have the final velocity to find the time so;

$\rm v^2=u^2+2gh\\\\ \rm v^2=8^2+2\times 9.81 \times 8.6 \\\\ \rm v= \sqrt{8^2+2\times 9.81 \times 8.6}\\\\\rm v=15.03 \;m/sec$

According to Newton's third equation of motion ;

$\rm v=u-gt \\\\ \rm t=\frac{v-u}{g} \\\\ \rm t=\frac{15.03-8}{9.81} \\\\ \rm t=0.717 sec.$

Hence the time would it take to reach the ground will be 0.717 sec.

To learn more about the velocity refer to the link ;

brainly.com/question/862972

3 0

2 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Which activity is best described as a scientific endeavor?

Kipish [7]

Answer:

Explanation:

BECAUSE TO DO THE TESTS YOU NEED TO DO THE SCIENTIFIC METHOD.

FOR EXAMPLE: OBSERVATIONS AND EXPERIMENTS TO OBTAIN RESULTS.

ANYWAY I LEAVE YOU THE LINK:

https://gscourses.thinkific.com

8 0

2 years ago

A red train travelling at 72 km/h and a green train travelling at 144 km/h are headed toward each

abruzzese [7]

Answer:

Collision will occur.

Speed of red train when they collide = 0 m/s.

Speed of green train when they collide = 10 m/s.

Explanation:

Speed of red train = 72 km/h = 20 m/s

Speed of green train = 144 km/h = 40 m/s.

Deceleration of trains = 1 m/s²

For red train:-

Equation of motion v = u + at

u = 20 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 20 - 1 x t

t = 20 s.

Equation of motion s = ut + 0.5at²

u = 20 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 20 x 20 - 0.5 x 1 x 20² = 200 m

So red train travel 200 m before coming to stop.

For green train:-

Equation of motion v = u + at

u = 40 m/s

v = 0 m/s

a = -1 m/s²

Substituting

0 = 40 - 1 x t

t = 40 s.

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 40 s

a = -1 m/s²

Substituting

s = 40 x 40 - 0.5 x 1 x 40² = 800 m

So green train travel 800 m before coming to stop.

Total distance traveled = 800 + 200 = 1000 m>950 m.

So both trains collide.

Distance traveled by green train when red train stops(t=20s)

Equation of motion s = ut + 0.5at²

u = 40 m/s

t = 20 s

a = -1 m/s²

Substituting

s = 40 x 20 - 0.5 x 1 x 20² = 600 m

Total distance after 20 s = 600 + 200 = 800 m< 950m . So they collide after red train stops.

Speed of red train when they collide = 0 m/s.

Distance traveled by green train when they collide = 950 - 200 = 750 m

Equation of motion v² = u² + 2as

u = 40 m/s

s= 750 m

a = -1 m/s²

Substituting

v² = 40² - 2 x 1 x 750 = 100

v = 10 m/s

Speed of green train when they collide = 10 m/s.

6 0

3 years ago

Think about Newton's 2nd Law, Force equals mass times acceleration. Liz puts a 1 kg weight and a 10 kg weight on identical wagon

Alexeev081 [22]

More force is needed for more mass. Therefore, if the mass is greater and the force is not enough then the object will less likely accelerate

5 0

2 years ago

Read 2 more answers