A student walks 3.0m [E] and then 4.0m [N] in 5.0s.

Neko [114]

Answer:

hey...................................

8 0

3 years ago

A race car accelerates on a straight road from rest to speed of 180 km/hr in 25s . Assuming uniform acceleration of the car thro

adelina 88 [10]

X-x0=0.5(v0+v)t ergo x=0.5(50)25=625m

3 0

4 years ago

Calculate the power of the elevator if it can raise 290 kg of mass 30 flights of stairs (300 m) in 17 seconds

Advocard [28]

Answer:

Power = 50204 [watts]

Explanation:

We know that the power is defined by the following expression:

Power = Work/time

where:

Power [watts]

time [seconds]

The work done will be the following:

Work = Force * distance [Joules]

Force[Newtons]

distance[meters]

Force = mass* gravity

Force=290 [kg]*9.81[m/s^2] =2844.9[N]

Work = 2844.9[N]*300[m] = 853470[J]

Therefore

Power = 853470 / 17 = 50204 [watts]

5 0

4 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

3 years ago

A boat moves initially at 10 m/s, then accelerates at 2 m/s^2 for 10 s. What is the velocity of the boat after 10 s?

bija089 [108]

Answer:

Explanation:

acceleration is the change in velocity of a body with respect to time

a = v-u/t

v is the final velocity

u is the initial velocity = 10m/s

t is the time taken = 10s

a is the acceleration = 2m/s²

substitute the values into the formula

2 = v-10/10

cross multiply

20 = v-10

v = 20+10

v = 30m/s

Hence the velocity of the body after 10s is 30m/s

6 0

4 years ago