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Lubov Fominskaja [6]
2 years ago
15

A car speeds up from 18.54 m/s to 29.52 m/s in 13.84 s. The acceleration of the car is:

Physics
2 answers:
valkas [14]2 years ago
5 0

Answer:

.7934m/s^{2}

Explanation:

Acceleration = change in velocity / change in time

A = 10.98m/s / 13.84s

A = .7934m/s^{2}

raketka [301]2 years ago
3 0

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

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True or False: The basketball should be dribbled below the waist.
zimovet [89]
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.
7 0
2 years ago
If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.
Schach [20]

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

6 0
3 years ago
Identify two everyday phenomena that exhibit diffraction of sound and explain how diffraction of sound applies. Identify two eve
Naya [18.7K]

Answer:

Explanation:

Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.

1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:

i. The voices of people talking outside a building can be heard by those inside.

ii. The sound from the horn of a car can be heard by people at certain distances away.

When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.

2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:

i. The shadow of objects which has the umbra and penumbra regions.

ii. The apparent color of the sky.

A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.

Some significant differences between diffraction of light and that of the sound are:

i. Diffraction of light is not as common as that of sound.

ii. Sound propagates through a wider region than light waves.

iii. Sounds are longitudinal waves, while lights are transverse waves.

7 0
3 years ago
Reactance Frequency Dependence: Sketch a graph of the frequency dependence of a resistor, capacitor, and inductor. RLC Circuit R
jolli1 [7]

Answer:

f=\frac{1}{2\pi \sqrt{LC}}

Explanation:

We know that impedance of a RLC circuit is given by Z=R+J(X_L-X_C)

So Z=\sqrt{R^2+(X_L-X_C)^2} here R is resistance X_L is inductive reactance and X_C is capacitive reactance

To minimize the impedance X_L-X_C should be zero we know that X_L=\omega L\ and \ X_C=\frac{1}{\omega C}

So \omega L-\frac{1}{\omega C}=0

\omega ^2=\frac{1}{LC}

\omega =\sqrt{\frac{1}{LC}}

We know that \omega =2\pi f

So \omega =2\pi f=\frac{1}{\sqrt{LC}}

f=\frac{1}{2\pi \sqrt{LC}}

Where f is resonance frequency  

8 0
2 years ago
Which pm the following are likely to form a covalent bond
Alecsey [184]

Answer:

magneisuim and gold

Explanation:

8 0
3 years ago
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