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3 years ago

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?

Physics

1 answer:

Helen [10]3 years ago

8 0

Answer:

the car have travelled 0.31 mile during that time

Explanation:

Applying the Equation of motion;

s = 0.5(u+v)t

Where;

s = distance travelled

u = initial speed = 0 mph

v = Final speed = 50 mph

t = time taken = 3/4 min = 3/4 ÷ 60 hours = 1/80 hour

Substituting the given values into the equation;

s = 0.5(0+50)×(1/80)

s = 0.3125 miles

s ~= 0.31 mile

the car have travelled 0.31 mile during that time

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

A large, simple pendulum is on display in the lobby of the United Nations building. If the pendulum is 18.5 m in length, what is

zhannawk [14.2K]

Answer:

t = 1,144 s

Explanation:

The simple pendulum consists of an inextensible string with a mass at the tip, the angular velocity of this is

w = √( L / g)

The angular velocity is related to the frequency and period

w = 2π f

f = 1 / T

w = 2π / T

Let's replace

2π / T = √ (L / g)

T = 2π √ (g / L)

Let's calculate

T = 2π √ (9.81 / 18.5)

T = 4,576 s

The definition of period in the time it takes the ball to come and go to a given point (a revolution) in our case we go from the end to the middle point that is a quarter of the path

t = T / 4

t = 4,576 / 4

t = 1,144 s

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3 years ago

Which of these black holes exerts the weakest tidal force on an object near its event horizon? Which of these black holes exerts

ratelena [41]

Answer:

THA ANSWER IS B

Explanation:

YEH ITS B

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3 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

Ulleksa [173]

<span>s= 0.5 at^2
</span>
0.5 x 9.81 x 3^2 = 44m

8 0

3 years ago

Read 2 more answers

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the

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<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal

it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
--------------------------------------...
let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
=====================
t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s

so it will able to cross</span>

8 0

3 years ago

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that​ time?

A car accelerates at a constant rate from 0 to 50 mph in three fourths min. How far does the car travel during that time?