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3 years ago

Three objects each with a mass of 10.0 kg are

Physics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

Answer: F net=0

Explanation:

Its in the picture.

Alenkasestr [34]3 years ago

7 0

Answer: The force is either attractive or repulsive between them creates a net force of '0' at the center

Explanation:

$F=G.m.m/dx^{2}$ The forces are equal to each other

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A proton moves in the negative x-direction through a uniform magnetic field in the negative y-direction what is the direction of

ANEK [815]

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.

<h3>What is the right-hand thumb rule?</h3>

Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.

A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.

The proton will be subject to a magnetic pull that is directed into the page.

Hence, option B is correct.

To learn more about the right-hand thumb rule refer to the link;

brainly.com/question/11521829

#SPJ1

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2 years ago

What is the speed of a bobsled whose distance-time graph indicates that it traveled 122m in 27s?

Stolb23 [73]

Speed = distance/time
speed= 122÷27=4.52m/s (3sf)

7 0

3 years ago

Amy is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward; and Carlos is driving his car at 15 m/s westw

Solnce55 [7]

Answer:

20 m/s westward

Explanation:

Taking eastward as positive direction, we have:

$v_B = +5 m/s$ is the velocity of Bill with respect to Amy (which is stationary)

$v_C = -15 m/s$ is the velocity of Carlos with respect to Amy

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

$v_B = +5 m/s$

Therefore, Carlos velocity in Bill's reference frame will be

$v_C' = v_C - v_B = -15 m/s - (+5 m/s) = -20 m/s$

and the direction will be westward (negative sign).

3 0

3 years ago

In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat

olya-2409 [2.1K]

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

$v_f = 0$

now we can use kinematics top find the acceleration of the bullet

$v_f^2 - v_i^2 = 2 a d$

so we have

$0 - 55^2 = 2(a)(1.34)$

$a = -1128.7 m/s^2$

now by Newton's II law we know that

$F = ma$

so we have

$F = (0.052)(-1128.7)$

$F = -58.7 N$

8 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago