Bryan is riding his scateboard at 2.0 m/s. He accelerates at 0.5 m/s for 4.0 seconds. Calculate his final velocity.

Answers

Answer 1
Answer:

Answer:

4 m/s^2

Explanation:

(0.5m/s * 4)+2 m/s


Related Questions

A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.
If the_____of a wave increases, its frequency must decrease.a. period b. energy c. amplitude d. velocity
A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s
A rocket is attached to a toy car that is confined to move in the x-direction ONLY. At time to = 0 s, the car is not moving but the rocket is lit, so the toy car accelerates in the +x-direction at 5.35 m/s2. At t; = 3.60 s, the rocket's fuel is used up, and the toy car begins to slow down at a rate of 1.95 m/s2 because of friction. A very particular physics professor wants the average velocity for the entire trip of the toy car to be +6.50 m/s. In order to make this happen, the physics professor plans to push the car (immediately after it comes to rest by friction) with a constant velocity for 4.50 sec. What displacement must the physics professor give the car (immediately after it comes to rest by friction) in order for its average velocity to be +6.50 m/s for its entire trip (measured from the time the rocket is lit to the time the physics professor stops pushing the car)?

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600  seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?

Answers

a). The magnitude along with the direction of the electric field releasing westward force of 4.80 × 10^(-17) N would be:

3 × 10^(-36) N/C is Eastward Direction

b). The magnitude along with the force of the direction that this field releases on proton would be:

4.8 × 10^(17) N in Eastward Direction

Electric Field

a). Given that,

Force =4.80 × 10^(-17) N

As we know,

Force= Charge × Electric Field

So,

∵ Electric Field= Force/Charge

= 4.8 × 10^(17))/(1.6 × 10^(-19))

= 3 × 10^(36)N/C

The direction of the field would be opposite i.e. Eastward direction due to the field carrying a -ve charge.

b). The magnitude carried by the force working on the proton would be the same with an opposite direction due to +ve charge.

∵ Force  =4.80 × 10^(-17) N in Eastward direction.

Learn more about "Magnitude" here:

brainly.com/question/9774180

Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

Two coils are wound around the same cylindrical form. When the current in the first coil is decreasing at a rate of -0.245 A/s , the induced emf in the second coil has a magnitude of 1.60×10−3 V . Part A What is the mutual inductance of the pair of coils? MM = nothing H Request Answer Part B If the second coil has 22 turns, what is the flux through each turn when the current in the first coil equals 1.25 A ? ΦΦ = nothing Wb

Answers

To solve this problem we need to use the emf equation, that is,

E=m(dI)/(dT)

Where E is the induced emf

I the current in the first coil

M the mutual inductance

Solving for a)

M=(E)/((dI)/(dT))\nM=(1.6*10^(-3))/(0.245)=6.53*10^(-3)H

Solving for b) we need the FLux through each turn, that is

\Phi=(MI)/(N)

Where N is the number of turns in the second coil

\Phi=(6.53*10^(-3)*1.25)/(22)=3.71*10^(-4)Wb

I'm a little bit unsure about this question.

Answers

Answer:

Option C. 4 Hz

Explanation:

To know the correct answer to the question given above, it is important we know the definition of frequency.

Frequency can simply be defined as the number of complete oscillations or circles made in one second.

Considering the diagram given above, the wave passes through the medium over a period of one second.

Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.

From the diagram given above,

The number of circles = 4

Thus,

The frequency is 4 Hz

An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?

Answers

Answer:

The S wave arrives 6 sec after the P wave.

Explanation:

Given that,

Distance of P = 45 km

Speed of p = 5000 m/s

Speed of S = 3000 m/s

We need to calculate the time by the P wave

Using formula of time

t = (D)/(v)

Where, D = distance

v = speed

t = time

Put the value in to the formula

t_(p) =(45*1000)/(5000)

t_(p) = 9\ sec

Now, time for s wave

t_(s)=(45000)/(3000)

t =15\ sec

The required time is

\Delta t=t_(s)-t_(p)

\Delta t=15-9

\Delta t =6\ sec

Hence, The S wave arrives 6 sec after the P wave.

A baseball is thrown horizontally with a velocity of 44 m/s. It travels a horizontal distance of 18m to the plate before it is caught. How long does the ball stay in the air?

A.0.41 sec
B.41 sec
C.4.1 sec
D.4 sec

Answers

A horizontal baseball pitch is launched at 44 m/s. The ball will stay for 4.1 sec (approx) in the air. Hence, option C is correct.

What is Velocity?

The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.

Its SI unit is represented as m/s, and it is a vector quantity, it means that it has both magnitude and direction.

According to the question, the given values are :

Initial Velocity, u = 44 m/s,

Distance travelled, s = 18 m and,

Final velocity, v = 0.

Use equation of motion :

v = u + at

0 = 44 + (-9.8)t

t = 44 / 9.8

t = 4.3 (approx)

Hence, the time for which the ball stay in the air is 4.1 sec (approx).

To get more information about velocity :

brainly.com/question/13372043

#SPJ2

Answer:

a 0.41

plug number into equation