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2 years ago

6

An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius

of the circle?

1 answer:

Gnom [1K]2 years ago

7 0

Answer:

r = mv^2/ Fc

Explanation:

Got it right

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Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro

pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0

3 years ago

Read 2 more answers

A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili

lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

* Bimetallic brass / steel tape

* Initial temperature, room temperature T = 20ºC

* Final temperature, boiling water = 100ºC

* initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

* thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

* perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass α_{brass} = 19 10⁻⁶ ºC⁻¹

Steel α_{steel} = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass L_{f \ brass} =θ r₁

steel L_{f \ steel} = θ r₂

we substitute in our equations

θ r₁ - L₀ = α_{brass} L₀ ΔT

θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

e = r₁ -r₂

θ = $Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}$

we calculate

θ = $22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}$

θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

tan θ = Δy / L₀

Δy = L₀ tan θ

Δy = 22.2 tan 0.155

Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0

2 years ago

Describe the difference between contact forces and field forces.

Salsk061 [2.6K]

Contact forces has to be touching for it to be an actual force. A field force does not have to be touching but it does have to be acting on particles at different positions in a space.

8 0

3 years ago

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio

lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

$W = F \times d = 2.62 N \times 100 m$

$W = 261.6 N\times m$

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

$0.5 m\times v2^2 = 0.5 m\ v1^2 - W$

Now solve for v2

$v2 = \sqrt{v1^2 - {\frac{2W}{M}}}$

$= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}$

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

$Ff = Us\times N$

Now solve for Us

$= \frac{Ff}{N}$

$= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}$

= 0.00494

4 0

3 years ago

A mass accelerates uniformly when the resultant force on it:

WARRIOR [948]

F = ma
So if you want the "a" to stay constant (as it is then uniform acceleration), the F must also be constant to apply a constant force.

2 is the answer

3 0

3 years ago