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3 years ago

A dropped ball accelerates as it falls. but a ball rolling on a horizontal surface has zero acceleration because _______.

Physics

2 answers:

svet-max [94.6K]3 years ago

5 0

Answer:

There is no force acting on it.

Explanation:

The dropping ball experiences the force of gravity on it that make sit accelerate towards the ground constantly, if the ball is rolling on a horizontal surface the effect of the gravity will just have an effect on the weight of the ball and the friction, but unless you are applying an external continuos force to that ball it would not experience acceleration.

nordsb [41]3 years ago

4 0

During a fall, the ball experiences an external force which is acted upon by gravity. This gravitation force induces a constant gravitational acceleration on the object equivalent to 9.81 meters per square second. Hence, the dropped ball is accelerating.

On the other hand, the ball does not experience external force hence the acceleration is zero.

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A car starts from rest and speeds up to 30 m/s in a total of 6 s.

sveta [45]

5 m/s
30 divided by 6 is 5

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2 years ago

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What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a

Elena L [17]

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

6 0

3 years ago

A 85 kg lineman tackles a 90 kg receiver. The receiver is running 5.8 m/s, and the lineman is moving 4.1 m/s, at a right angle t

weeeeeb [17]

Answer:

3.59 m/s

Explanation:

We are given that

Mass of lineman,m=85 kg

Mass of receiver,m'=90 kg

Speed of receiver,v'=5.8 m/s

Speed of lineman,v=4.1 m/s

$\theta=90^{\circ}$

We have to find the their velocity immediately after the tackle.

Initial momentum, $P_i=\sqrt{p^2_1+p^2_2}=\sqrt{(85\times 4.1)^2+(90\times 5.8)^2}=627.6 kgm/s$

According to law of conservation of momentum

Initial momentum=Final momentum= $(m+m')V$

$627.6=(85+90)V=175V$

$V=\frac{627.6}{175}=3.59 m/s$

3 0

3 years ago

Consider the following waves representing electromagnetic radiation: An illustration shows two waves representing electromagneti

Murljashka [212]

Answer:

a) red wave hs a longer wavelength than the green wave

b)f = 1.875 10¹¹ Hz , f_green = 5.45 10¹⁴Hz

c) E = 1.24 10⁻²² J , E_green = 3.6 10⁻¹⁹ J

d) The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

Explanation:

a) The green wave are on the left in the electromagnetic spectrum so the red wave has a longer wavelength than the green wave

The green wavelength is in the range of 550 10⁻⁹ m

The speed of the wave is

c = λ f

f = c /λ

b) The frequency of the red wave is

f = 3 10⁸ / 1.6 10⁻³

f = 1.875 10¹¹ Hz

For the green wave

f_green = 3 10⁸/550 10⁻⁹

f_green = 5.45 10¹⁴Hz

c) The photon energy is given by the Planck equation

E = h f

E = 6.63 10⁻³⁴ 1.875 10¹¹

E = 1.24 10⁻²² J

For the green wave

E_green = 6.63 10⁻³⁴ 5.45 10¹⁴

E_green = 3.6 10⁻¹⁹ J

d) The speed of electromagnetic waves is constant and has a value of 3 108 m / s

The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

6 0

2 years ago

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

3 0

3 years ago