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3 years ago

How do you solve this question. I found the first part but don't know how to find distance

Physics

2 answers:

Levart [38]3 years ago

7 0

The distance it traveled during that time is

(average speed) x (length of time) .

The average speed is

(1/2) (starting speed) + (ending speed) .

They gave you the starting speed.
You calculated the ending speed.
So you can calculate the average speed.

They gave you the length of time.
So you can calculate the distance.

Tpy6a [65]3 years ago

5 0

Why don't you try using the following kinematics formula

Vf^2 = Vi^2 + 2ad

Then solve for d, distance.

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Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo

Tems11 [23]

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

3 0

4 years ago

If 1.34 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved throu

ioda

Given :

Number of operations move through a pocket calculator during a full day's operation , $n=1.34 \times 10^{20}$ .

To Find :

How many coulombs of charge moved through it .

Solution :

We know , charge in one electron is :

$e^-=-1.6\times 10^{-19}\ coulombs$

So , charge on n electron is :

$C=e^-\times n\\C=-1.6\times 10^{-19}\times 1.34\times 10^{20} \ C\\C=-21.44\ C$

Therefore , -21.44 coulombs of charge is moved through it .

Hence , this is the required solution .

3 0

3 years ago

A boat moves 60 kilometers east from point A to point B. There, it reverses direction and travels another 45 kilometers toward p

ElenaW [278]

Total distance = 15 kilometers and total displacement = 110 kilometers----- 60 kilometers - 45 kilometers = 15 kilometers 60 + 45 kilometers = 110 kilometers :)))))

3 0

3 years ago

A cart with mass 2.0 kg moving on a frictionless linear air track at an initial speed of 1.0 m/s undergoes an elastic collision

Katen [24]

Answer:

a) P=0.8 Kg*m/s b) K=0.6 N c)P/K=MV/(1/2*MV²) d) V2f=1.5 m/s e) M2=0.53 Kg

Explanation:

During an elastic collision between 2 bodies, the momentum P is the same before and after the collision

For this case:

Before the collision:

M₁= mass of first car= 2 Kg

V₁= initial speed of the first car = 1 m/s

M₂= mass of the second car

V₂= initial speed of the second car = 0 m/s (as it is stationary)

After the collision:

V₁f= final speed of the first car after the collision= 0.6 m/s

V₂f= final speed of the second car after the collision

As momentum is the same after and before:

M₁V₁ + M₂V₂ = M₁V₁f + M₂V₂f consider that term M₂V₂=0 as V₂=0

Then, momentum for car N° 2 after the collision is: P₂= M₂V₂f and replacing from the above equation: P₂= M₁V₁ – M₁V₁f = M₁(V₁ – V₁f) = 2 Kg*(1m/s – 0.6m/s) = 0.8 Kg*m/s

As the kinetic energy “K” is also conservative:

½*M₁V₁² + ½*M₂V₂² = ½*M₁V₂f² + ½*M₂V₂f² Where ½*M₂V₂²=0

Then: K₂= ½*M₂V₂f² = ½*M₁(V₁² – V₁f²) = 0.64 N

Finally, to obtain M₂ and V₂f:

P₂=M₂V₂f and K₂=1/2*M₂V₂f2²

P₂/K₂= (M₂V₂f)/(1/2*M₂V₂f) =2/V₂f

V₂f= 2*K₂/P₂=1.5 m/s and M₂=P₂/V₂f=0.53 Kg

8 0

3 years ago

a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig

suter [353]

The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

$2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}$

Then, the final velocity will be

$v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s$

Thus, the final velocity of the truck is found as 146.969 m/s.

8 0

3 years ago