4.6 billion years ago, the Earth is created from what

Sarah, who has a mass of 55 kg, is riding in a car at 20 m/s. She sees a cat crossing the street and slams on the brakes! Her se

avanturin [10]

Answer:

-2200 N

Explanation:

The change in momentum of Sarah is equal to the impulse, which is the product between the force exerted by the seatbelt on Sarah and the time during which the force is applied:

$\Delta p=I\\m \Delta v = F \Delta t$

where

m is the mass

$\Delta v$ is the change in velocity

F is the average force

$\Delta t$ is the duration of the collision

In this problem:, we have:

m = 55 kg is Sarah's mass

$\Delta v = 0-20 = -20 m/s$ is the change in velocity

$\Delta t = 0.5 s$ is the duration of the collision

Solving for F, we find the force exerted by the seatbelt on Sarah:

$F=\frac{m\Delta v}{\Delta t}=\frac{(55)(-20)}{0.5}=-2200 N$

Where the negative sign indicates that the direction of the force is opposite to that of Sarah's initial velocity.

5 0

3 years ago

Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45

victus00 [196]

<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>

4 0

2 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

7 0

2 years ago

A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio

Leya [2.2K]

Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

3 0

3 years ago

Read 2 more answers

It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.

Cerrena [4.2K]

D Because If Your Going To Have A Contest Its Ganna Have To Be The Same Objectives For Both Contenders

3 0

2 years ago

Read 2 more answers