Ask question

Ask question

All categories

SVETLANKA909090 [29]

2 years ago

9

A crime suspect fled the scene when police officers entered the building. How far could he run in 10 minutes if he can run 5 mil

es per hour?

1 answer:

emmainna [20.7K]2 years ago

3 0

(5 mi/hr) x (1hr/60min) x (10min) = 5 x 10 / 60 = <em>5/6 mile</em>

(5/6 mile) x (1,760 yd/mile) = <em>1,466 and 2/3 yards</em>

You might be interested in

A mass m is attached to an ideal massless spring. When this system is set in motion with amplitude a, it has a period t. What is

Luden [163]

The period will be the same if the amplitude of the motion is increased to 2a

What is an Amplitude?

Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

Here,

mass m is attached to the spring.

mass attached = m

time period = t

We know that,

The time period for the spring is calculated with the equation:

$T = 2\pi \sqrt{\frac{m}{k} }$

Where k is the spring constant

Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.

From the equation, we can say,

Time period of the spring is independent of the amplitude.

Hence,

Increasing the amplitude does not affect the period of the mass and spring system.

Learn more about time period here:

<u>brainly.com/question/13834772</u>

#SPJ4

7 0

1 year ago

A 1600 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2000 kg SUV traveling from ea

skad [1K]

Answer:

v = 11.1m/s

Vx = 7.19m/s

Vy = 8.45m/s

V(sedan) = 30.4m/s

V(suv) = 25.9m/s

Explanation:

4 0

2 years ago

An electric current in a metal consists of moving

Alik [6]

The answer would be (A) Protons

6 0

2 years ago

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

1.

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

2.

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

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

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

2 years ago

A pitcher throws a baseball at 45 m/s. The baseball has a mass of 400 grams. Disregarding air resistance, the baseball's momentu

enyata [817]

So it would be 8.90 m/s kg*m/s

5 0

2 years ago