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3 years ago

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

Physics

2 answers:

ycow [4]3 years ago

7 0

Answer:

The speed of the ball is 10 meters per second.

Explanation:

The fact that ball rotates uniformly means that speed of the ball is constant and resulting acceleration is centripetal, which means that external horizontal force exerted on the ball goes towards the center of rotation. We can estimate the speed of the ball by this equation of equilibrium:

$\Sigma F_{r} = F = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

Where:

$F$ - External force exerted on the ball, measured in newtons.

$m$ - Mass of the ball, measured in kilograms.

$v$ - Linear rotation speed, measured in meters per second.

$R$ - Radius of rotation, measured in meters.

We clear the linear speed in (Eq. 1):

$v^{2} = \frac{F\cdot R}{m}$

$v = \sqrt{\frac{F\cdot R}{m} }$

If we know that $F = 400\,N$ , $m = 2\,kg$ and $R = 0.50\,m$ , the speed of the ball is:

$v =\sqrt{\frac{(400\,N)\cdot (0.50\,m)}{2\,kg} }$

$v = 10\,\frac{m}{s}$

The speed of the ball is 10 meters per second.

frutty [35]3 years ago

6 0

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

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3 years ago

One difference between a solar flare and a CME is that a solar flare is composed of ___________, while a CME is composed of ____

JulsSmile [24]

Answer:

magnetic energy (proton) and magnetic plasma.

Explanation:

The solar fare consists of bright light that occurs in various wavelengths and is observed at the surface.

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The Solar flares represent an atmospheric disturbance and plasms are the medium for the growth and development of solar flare and lead to solar activity.

6 0

3 years ago

A shell is fired with a horizontal velocity in the positive x direction from the top of an 80 m high cliff. The shell strikes th

Sonbull [250]

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

7 0

3 years ago

Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu

ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

= 223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0

3 years ago

Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 9.0 km/h due east. Runner B is initia

sergejj [24]

Answer:

0.176m from the flagpole, westward.

Explanation:

Let the Eastward be the positive direction. So initially runner A is at position -6km, running with velocity of 9km/h while runner B is at position 5km running at a velocity of -8km/h. We can conduct the following equation for their distances over the same time t

$s_A = -6 + 9t$