Someone please help me

can somebody help me answer this question? A car went from 110 m/s to 80 m/s in 20 seconds. What was the acceleration of the car

gogolik [260]

Answer:

-1.5m/s²

Explanation:

Acceleration can be thought of as [Change in Velocity]/[Change in time]. To find these changes, you simply subtract the initial quantity from the final quantity.

So for this question you have:

V_i = 110m/s
V_f = 80m/s
t_i = 0s
t_f = 20s

which means that the acceleration = (80-110)/(20-0)[m/s²] = (-30/20)m/s² = -1.5m/s²

4 0

2 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

If a 40,000g cannon ball is 35m above the Earth's surface, how much potential energy does the cannon

mihalych1998 [28]

Answer: 21

Explanation:

because 9 + 10 = 21

8 0

3 years ago

The driver of a car going 96.0 km/h suddenly sees the lights of a barrier 39.0 m ahead. It takes the driver 0.75 s to apply the

arsen [322]

Answer:

21.42m/S

Explanation:

Hello!

To solve this problem we must perform the following steps.

1. Find the distance traveled from until the driver reacts, this is achieved using the equation for constant speed movement.

X1=VT

where

x= distance traveled

v=initial speed

T=time=0.75s

X1=0.75Vo

we must take into account that the total distance is the sum of the distance at which the pilot reacts (x1) and when it starts to decelerate (x2)

39=0.75Vo+X2

X2=39-0.75Vo

2. Now we use the equation that defines a movement with constant acceleration.

$Vo =\sqrt{Vf^2 - 2(a)(x2)}$

where

Vf = final speed=0m/s

Vo = Initial speed

A = acceleration =-10m/s2

X2 = displacement

now we use the ecuation of step 1

$Vo =\sqrt{Vf^2 - 2(a)(39-0.75Vo)}$

solving

$Vo =\sqrt{0 - 2(-10))(39-0.75Vo)}\\Vo^2=780-15Vo\\Vo^2+15Vo-780=0$

Now we solve the quadratic equation and find the value of Vo

the solutions are 21.42m/S, -36.41m/S

as the speed must be positive we conclude that the answer is 21.42m/S

6 0

3 years ago

A 45.2-kg wagon is towed up a hill inclined at 18.5∘ with respect to the horizontal. The tow rope is parallel to the incline and

drek231 [11]

Answer:

The speed is 29.9 m/s

Explanation:

The force created from gravity due to the wagon mass is:

$F=m*g*sin(18.5)\\F=45.2*9.8*sin(18.5)\\F=140.55$

140.55 N pull the wagon down. Two parallel rope with tension of 191N creates 382 N on the wagon. Therefore:

$T_{total}=382-140.55=241.45$

241.45 N force is pulling up the wagon. Then we need to find the acceleration of the wagon under this force:

$F=m*a\\241.45=45.2*a\\a=5.34$

acceleration is 5.34 m/s^2. The distance is multiplication of acceleration and square of the time.

$x=1/2*a*t^2\\83.8=0.5*5.34*t^2\\t=5.6$

After 5.6 second the wagon will ride 83.8 m up to hill. And the speed of wagon at that point is:

$v=a*t\\v=29.9$

7 0

3 years ago