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2 years ago

11

A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t

he cyclist's initial velocity? 4 m/s 12 m/s 18 m/s 20 m/s

2 answers:

Mekhanik [1.2K]2 years ago

7 0

Answer:

4m/s

Explanation:

e2020

Blababa [14]2 years ago

3 0

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

$a=\frac{\Delta v}{t}$

Where

a is the acceleration

v is velocity

t is time

$\Delta$ is "change in"

For this problem , we are given

a = 1.2

t = 10

Putting into formula, we get:

$a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12$

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

$\Delta v = Final \ Velocity - Initial \ Velocity$

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

$\Delta v = Final \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4$

hence,

Initial Velocity is 4 m/s

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

2 years ago

A boy is pulling a load of 150N with a string inclined at an angle 30 to the horizontal if the tension of string is 105N the for

Lorico [155]

The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>

Why?

Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.

We can calculate the vertical force using the following formula:

$VerticalForce=Force*Sin(30\° )=(BoysForce-StringForce)*\frac{1}{2}\\\\VerticalForce=(150N-105N)*\frac{1}{2}=VerticalForce=45N*\frac{1}{2}=22.5N$

Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>

Have a nice day!

8 0

2 years ago

I pull the throttle in my racing plane at a = 12.0 m/s2. I was originally flying at v = 100. m/s. Where am I when t = 2.0s, t =

Helen [10]

Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)

• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)

•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)

7 0

2 years ago

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

Vikki [24]

acceleration of rocket is given here as

$a_y = 2.60* t$

now we know that

$\frac{dv}{dt} = 2.60t$

now integrating both sides

$\int dv = \int 2.60t dt$

$v = 2.60\frac{t^2}{2}$

$v = 1.30 t^2$

here since its given that rocket will accelerate for t = 10 s

so here we have

$v = 1.30 * 10^2$

$v = 130 m/s$

so after t = 10 s the speed of rocket will be 130 m/s upwards

5 0

2 years ago

Match the measuring instrument with the appropriate statement.

yaroslaw [1]

<u>Answer:</u>

1A, 2E, 3C, 4F, 5B, 6D

<u>Explanation:</u>

Tilt meter is a an equipment used to measure very small changes from vertical level. So it is a correct match for blank space in statement A.

Richter Scale is a scale of measurement of earthquake strength, So it is a correct match for blank space in statement E.

Mercalli Intensity Scale measures effect of an earthquake, like damage caused.So it is a correct match for blank space in statement C.

Seismograph is earthquake wave measuring instrument. So it is a correct match for blank space in statement F.

Correlation Spectrometer (C.O.S.P.E.C.) is used to measure sulfur dioxide content in smoke. So it is a correct match for blank space in statement B.

Moment Magnitude Scale is an earthquake measuring scale for great earthquakes. So it is a correct match for blank space in statement D.

7 0

2 years ago