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2 years ago

A car is moving at 14 m/s. After 30 s, its speed inncreased to 20 m/s. What is the acceleration over time

Physics

1 answer:

Juli2301 [7.4K]2 years ago

7 0

Answer:

0.2 m/s^2

Explanation:

initial speed 14m/s

final speed 20m/s

acceleration:

(20m/s - 14m /s) /30s = (6m/s)/30s = 0.2 m/s^2

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Which statement describes a characteristic of an experimental design that

Anarel [89]

Answer:

Explanation:

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3 years ago

A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil

miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0

3 years ago

Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg

muminat

Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

$V_{T}$ = V₁ +V₂ +V₃

This voltage may also be calculated using capacitance and charge;

V = Q/ C

$V_{T}$ = V₁ +V₂ +V₃

Provided that the total charge is 'q', hence the total voltage can be expressed as:

$V_{T}$ = (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)

Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.

Learn more about the capacitor here:

brainly.com/question/17176550

#SPJ4

7 0

1 year ago

A physical change is a change to the physical properties of a substance. This type of change does

LiRa [457]

Answer:

False

Explanation:

5 0

3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago