Ask question

Ask question

All categories

3 years ago

11

One car drives north at 60 mph. Another car leaces 30 minutes later and drives south at 50 mph. How long will it take for the tw

o cars to be 360 miles apart?

1 answer:

Softa [21]3 years ago

8 0

After the second car starts out,

D = 30 + 110 H

360 = 30 + 110 H

330 = 110 H

H = 330/110

H = 3 hours after the second car leaves.

That's 3.5 hours after the first car leaves.

You might be interested in

Alguém tem a resposta ?

Maru [420]

The answer is D or C

3 0

3 years ago

An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded

Alik [6]

Answer:

F = 1.68 N

Explanation:

Let's solve this exercise in parts.

Let's use the concept of conservation of the mechanical nerve

initial

Em₀ = 500 J

The energy is totally kinetic

Em₀ = K = ½ m v₀²

v₀ = $\sqrt{\frac{2 Em_{o} }{m} }$

v₀ = √ (2 500/32)

v₀ = 5.59 m / s

now with kinematics we can find a space

v² = v₀² - 2 a x

the negative sign is because the body is stopping

a = $( \frac{v_{o}^{2} - v^{2} }{2x} )$

let's calculate

a = (5.59² - 5.1²) / 2 50

a = 0.0524 m / s²

Finally let's use Newton's second law

F = ma

F = 32 0.0524

F = 1.68 N

7 0

3 years ago

A penny is made of the element zinc

AnnyKZ [126]

Answer:

Explanation: what is the question?

5 0

3 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago

A trombone has a variable length. When a musician blows into the mouthpiece and causes air in the tube of the horn to vibrate, t

lana66690 [7]

Answer:

The frequency increases with a shorter horn <em>(Option B)</em>.

Explanation:

The length of the horn determines the distance along which the wave travels; simply called the wavelength. Therefore, a short horn tube will produce a short wavelength and vice versa.

Sound waves have various characteristics that define pitches in musical instruments and these characteristics are interdependent on each other.

in this case, the frequency and the frequency and the wavelength are related.

The relationship between the wavelength and its frequency is given as:

<em> </em><em>c = f λ </em><em> </em>

<em>where 'c' is the speed of sound through the instrument; 'f ' is the frequency and 'λ' is the wavelength.</em>

Let's assume that the speed at which the musician blows air into the mouthpiece remains constant, an increase in wavelength will cause a decrease in frequency. Conversely, as the tube of the horn becomes shorter the frequency increases.

6 0

3 years ago