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MAVERICK [17]
3 years ago
15

A forklift raises a 72-kg crate onto a shelf 3.6 meters above the floor. How much gravitational potential energy does the crate

have relative to the floor?​
Physics
1 answer:
Mademuasel [1]3 years ago
8 0

Answer:

<h2>2540.16 J</h2>

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 72 × 9.8 × 3.6 = 2540.16

We have the final answer as

<h3>2540.16 J</h3>

Hope this helps you

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A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when
choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

F_t - mg = ma

here since speed is constant or it is at rest

so we will have

a = 0

F_t = mg

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

F_t - mg = ma

F_t = mg + ma

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

F_t - mg = -ma

F_t = mg - ma

so tension force is less than the weight of crate if it is accelerating downwards

4 0
3 years ago
At rest, hydrogen has a spectral line at 116 nm. if this line is observed at 107 nm for the star sirius, how fast is sirius movi
Citrus2011 [14]

2.3275862×10¹²km/s fast is sirius moving in km/s.

<h3>Briefing:</h3>

Hydrogen has a spectral line at = 116nm=116×10⁻⁹m

Line is observed at = 107 nm=107×10⁻⁹m

Now, from the Hubble's law

V=(\Delta \lambda / \lambda)×C

Where,

v is the velocity

Δλ = Change in wavelength = 116 - 107= 9nm=9×10⁻⁹m

λ = Actual wavelength=116nm=116×10⁻⁹m

C is the speed of the light=3×10⁸m/s

on substituting the respective values, we get

V=(9/116)×3×10⁸=23275862.069×10⁵m/s

V=2.3275862×10¹²km/s.

<h3>What is the wavelength?</h3>

A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. Typically, in wireless systems, this length is specified in meters (m), centimeters (cm), or millimeters (mm).

To know more about Wavelength visit:

brainly.com/question/13533093

#SPJ4

4 0
1 year ago
The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^
allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

   F = G  m1 m2 / r^2    

         G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11   *   4.41 x 10^5  * 5.97 x 10^24  / ( 6.78x 10^6)^2

 = 3820885 .3 N

6 0
2 years ago
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
f=- \frac{36}{7} cm=-5.1 cm

2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
r=2f=2 \cdot 5.1 cm=10.2 cm
3 0
3 years ago
How can two strengths be added together
Ket [755]
By using mind and dna
6 0
3 years ago
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