Ask question

Ask question

All categories

2 years ago

10

A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s2. What is the speed of the car after it has trav

eled 55 m?

1 answer:

Vedmedyk [2.9K]2 years ago

4 0

Answer:

The speed is =15.91ms−1. The time is =6.92s

Explanation:

You might be interested in

uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou

Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0

3 years ago

Which was least likely to have been a component of Earth’s atmosphere before life began?

Otrada [13]

The answer to this question is a) sulfur

4 0

3 years ago

Read 2 more answers

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia

masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

$V(t) = \epsilon. e^{-t.\frac{L}{R} }$

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

$0.8*19 = 19. e^{-4.\frac{22}{R} }$

$0.8 = e^{-\frac{88}{R} }$

$ln(0.8) = ln(e^{-\frac{88}{R} })$

$ln(0.8) = -\frac{88}{R}$

$R = -\frac{88}{ln(0.8)}$

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

3 years ago

What is the least amount of time required for a given point on this wave to move from y = 0 to y = 12cm?

ale4655 [162]

Time t = ?

<span>When wave is moving from y = 0 to y =12 cm</span>

By using the formula,

y = 15cos [(π/12) t)] = 0,

cos [(π/12) t)] = 0 = cos (π/2), so,

(π/12)t = π/2,

t = (π/2) (12/π)

t = 12/2

<span>t = 6 sec</span>

<span>so 6 sec is the least amount of time required</span>

4 0

3 years ago