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3 years ago

9

The net force acting on the ball below is ___ N

1 answer:

yKpoI14uk [10]3 years ago

8 0

The net force applied to the object equals the mass of the object multiplied by the amount of its acceleration." The net force acting on the soccer ball is equal to the mass of the soccer ball multiplied by its change in velocity each second (its acceleration).

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Two timpani (tunable drums) are played at the same time. One is correctly tuned so that when it is struck, sound is produced tha

yKpoI14uk [10]

Answer:

Explanation:

Given

Wavelength of first timpani $\lambda _1=2.2\ m$

Frequency corresponding to this Wavelength

$f_1=\frac{v}{\lambda _1}$

where $v=velocity\ of\ sound (343 m/s)$

$f_1=\frac{343}{2.2}$

$f_1=155.9\ Hz$

Wavelength of Second timpani $\lambda _2=2.1\ m$

Frequency corresponding to this Wavelength

$f_2=\frac{v}{\lambda _2}$

$f_2=\frac{343}{2.1}=163.33\ Hz$

Beat frequency $=f_2-f_1$

Beat frequency $=163.33-155.9=7.43$

so approximately 7 beats per second

4 0

4 years ago

Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predo

Zina [86]

Answer:

the answer is option A = photoelectric effect

Explanation:

If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.

6 0

3 years ago

How much energy is needed to melt 600 g of ice at 0 degrees C?

Keith_Richards [23]

Answer:

you need 300 butter

Explanation:

fadf

7 0

3 years ago

A car accelerates while trying to merge onto the freeway. Its speed goes from 0 km/h to 70km/h in 10 seconds. What is its accele

atroni [7]

Answer:

1.944m_s²

Explanation:

First convert speed from km/h to m/s the use the formula a=v-u

t

8 0

3 years ago

Two Charged spheres of 6.4 nC charge are 16 cm apart. What is the magnitude of the field in middle? ( n = 10^-9, 100 cm = 1 m)

emmainna [20.7K]

Answer:

zero

Explanation:

q = 6.4 nC = 6.4 x 1 0^-9 C

d = 16 cm = 0.16 m

r = 16 / 2 = 8 cm = 0.08 m

Electric field at P due to the charge placed at A

Ea = k q / r^2

Ea = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards right

Ea = 9000 Towards right

Electric field at P due to the charge placed at B

Eb = k q / r^2

Eb = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards left

Eb = 9000 Towards left

The magnitude of electric field is same but teh direction is opposite, so the resultant electric field at P is zero.

3 0

3 years ago