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Tanzania [10]
3 years ago
11

A roller coaster car starts at 6 m/s at the top of the hill at the 3 seconds latter at the bottom of the hill the roller coaster

car is going 72 m/s, what was the cars acceleration?
Physics
1 answer:
forsale [732]3 years ago
7 0

Answer:

22m/s

Explanation:

Step one:

Given data

Initial velocity of roller coster u= 6 m/s

Final velocity v= 72m/s

time t= 3 seconds

Required

We want to solve for the acceleration of the roller coster

Step two:

From the formula

acceleration= v-u/t

substitute

a= 72-6/3

a=66 /3

a= 22 m/s

The cars acceleration 22m/s

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A ball thrown straight up into the air is found to be moving at 6.79 m/s after falling 1.87 m below its release point. Find the
MaRussiya [10]

Answer:

3.07 m/s

Explanation:

We know that from kinematics equation

v^{2}=u^{2}+2as and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity

Making u the subject then

u=\sqrt {v^{2}-2gs}

Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then

u=\sqrt {6.79^{2}-(2\times 9.81\times 1.87)}=3.068338313 m/s\approx 3.07 m/s

5 0
4 years ago
A cello string vibrates in its fundamental mode with a frequency of 335 1/s. The vibrating segment is 28.5 cm long and has a mas
Inga [223]

Answer:

The tension in string is found to be 188.06 N

Explanation:

For the vibrating string the fundamental frequency is given as:

f1 = v/2L

where,

f1 = fundamental frequency = 335 Hz

v = speed of wave

L = length of string = 28.5 cm = 0.285 m

Therefore,

v = f1 2L

v = (335 Hz)(2)(0.285)

v = 190.95 m/s

Now, for the tension:

v = √T/μ

v² = T/μ

T = v² μ

where,

T = Tension

v = speed = 190.95 m/s

μ = linear mass density of string = mass/L = 0.00147 kg/0.285 m = 5.15 x 10^-3 kg/m

Therefore,

T = (190.95 m/s)²(5.15 x 10^-3 kg/m)

<u>T = 188.06 N</u>

4 0
3 years ago
Read 2 more answers
In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ
Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J

8 0
3 years ago
A 5000 kg open train car is rolling on frictionless rails at 22 m/s when it starts pouring rain. A few minutes later, the car’s
Phantasy [73]

Answer:

500 kg

Explanation:

It is given that,

The mass of a open train car, M = 5000 kg

Speed of open train car, V = 22 m/s

A few minutes later, the car’s speed is 20 m/s

We need to find the mass of water collected in the car. It is based on the conservation of momentum as follows :

initial momentum = final momentum

Let m is final mass

MV=mv

m=\dfrac{MV}{v}\\\\m=\dfrac{5000\times 22}{20}\\\\=5500\ kg

Water collected = After mass of train - before mass of train

= 5500 - 5000

= 500 kg

So, 500 kg of water has collected in the car.

3 0
3 years ago
If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w
DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

d = 2\pi r

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

d = 2\pi r = 2\pi*6371 km = 40030.2 km

Now, if we divide the above distance by the speed of the car we can find the time:

t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d

Therefore, Scobie will take 10 days to drive around Earth's equator.

     

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0
3 years ago
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