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Oksanka [162]
3 years ago
9

À A car moves with an initial velocity of 18 m/s due north. Find the velocity of the car after 7 Os if

Physics
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

(a) v_f=28.5m/s

(b) v_f=7.5m/s

Explanation:

Hello.

(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:

v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s

(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:

v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s

Best regards.

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A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w
user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

p\propto T

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

\frac{p_1}{T_1}=\frac{p_2}{T_2}

where in this problem:

p_1=0.981 atm is the initial pressure of the gas

T_1=65^{\circ}+273=338 K is the initial absolute temperature of the gas

T_2=41^{\circ}+273=314 K is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm

3 0
3 years ago
The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r
Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity.  Therefore:

marble > manhole cover > basketball > wedding ring

7 0
3 years ago
Radio waves travel at a speed of 300,000,000 m/s. WFNX broadcasts radio waves at a
yKpoI14uk [10]

Answer:

Wavelength, \lambda=2.94\ m

Explanation:

It is given that,

Speed of radio waves is v=3\times 10^8\ m/s

Frequency of radio waves is f = 101,700,000 Hz

We need to find the wavelength of WFNX’s radio waves. The relation between wavelength, frequency and speed of a wave is given by :

v=f\lambda

\lambda is wavelength

\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{3\times 10^8}{101,700,000}\\\\\lambda=2.94\ m

So, the wavelength of WFNX’s radio waves is 2.94 m.

3 0
3 years ago
1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una
romanna [79]

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

6 0
3 years ago
Which of the following are electromagnetic waves?
olchik [2.2K]

Answer:

Microwaves

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5 0
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