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marishachu [46]
3 years ago
14

A new piece of exercise equipment has been added to your gym, and you try it out. To use this machine, you lie horizontal on a m

at with your feet against a platform perpendicular to the mat. The platform is held by a stiff spring that is compressed when the platform moves. Your workout invovles compressing the spring by pushing on the platform with your feet. After testing compressions of the equipment, you determine that you must do 85.0 J of work to compress the spring a distance 0.170 m from its uncompressed length.What magnitude of force must you apply to hold the platform stationary at the final distance given above
Physics
1 answer:
Shalnov [3]3 years ago
6 0

Answer:

500 N

Explanation:

Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)

Since W = 85.0 J and we need to find F,

F = W/x

= 85.0 J/0.170 m

= 500 N

So, the  magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.

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What is a type of force that results from turning a corner on a bicycle? ​
Romashka-Z-Leto [24]

This force is centripetal force. it is the force that will hold the rider in the circular path and prevent him from skidding off. It is equal to the friction between the tyres and the road if no banking has been done on the road at the corner.

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3 years ago
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A 3.50 kg block is pushed along a horizontal floor by a force of magnitude 15.0 N at an angle θ = 30.0° with the horizontal. The
Crank

Answer :

The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

Explanation :

Given that,

Mass of block = 3.50

Angle = 30°

Force = 15.0 N

Coefficient of kinetic friction = 0.250

We need to calculate the frictional force

Using formula of frictional force

F_{k}=\mu N

F_{k}=\mu (F\sin\theta+mg)

F_{k}=0.250\times(15\times\sin30^{\circ}+3.50\times9.8)

F_{k}=0.250\times41.8

F_{k}=10.45\ N

(II). We need to calculate the block's acceleration

Using newton's second law of motion

F=ma

a=\dfrac{F}{m}

a=\dfrac{F\cos\theta-F_{k}}{m}

a=\dfrac{15.0\cos30^{\circ}-10.45}{3.50}

a=0.73\ m/s^2

Hence, The frictional force on the block from the floor and the  block's acceleration are 10.45 N and 0.73 m/s².

8 0
3 years ago
A rifle bullet with a mass of 11.5 g traveling toward the right at 251 m/s strikes a large bag of sand and penetrates it to a de
BabaBlast [244]

To look for the acceleration, it will come from:

vf^2=v0^2+2ad 
where:
vf = final velocity = 0 
v0 = initial velocity =251 m/s 
a = acceleration 
d= distance traveled = 0.237 m 

0=251^2+2a(0.237 ) 
a= -251 ^2 / (2*0.237) =-132 913.502 m/s/s 

we find the force from: 

F = ma = 0.0115kg*(-1.32x10^5m/s/s) = -1518 N 

the negative sign shows that the force is in the direction contradictory the bullet's motion

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3 years ago
A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

E=\frac{hc}{\lambda} is the energy of the incident photon, with

h = Planck constant

c = speed of light

\lambda = wavelength

\phi = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

eV=K

So we can rewrite (1) as

E=\phi + eV

where we have:

\lambda=200 nm = 2\cdot 10^{-7} m

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV

And now we can solve eq.(1) to find the work function of the metal:

\phi = E-eV=6.22 eV-1.93 eV=4.29 eV

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so \phi is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0
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A box, initially at rest, has 36.7 N of force exerted on it for 2.81 s. If the box has a mass of 7.41 kg, what was its velocity
ElenaW [278]

Answer:

13.91 m/s

Explanation:

First we need to find the acceleration:

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Then we find the velocity:

Velocity = Acceleration * Time

Velocity = 4.95 m/s² * 2.81 s

Velocity = 13.91 m/s (rounded to two decimal places)

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2 years ago
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