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nataly862011 [7]

2 years ago

15

The rate of change of momentum of a body free falling under gravity is equal to its? A. Velocity B. kinetic energy C. power D. w

eight

2 answers:

jolli1 [7]2 years ago

7 0

Answer:

weight

Explanation:

this is because of the lighter the object the faster it will be affected

ludmilkaskok [199]2 years ago

4 0

Your correct answer would be weight.

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1. You drop a rock off the top of a building. It takes 6,0 s. hit the ground. How tall is the building?

skelet666 [1.2K]

Not very y’all..........

5 0

3 years ago

Read 2 more answers

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur

sp2606 [1]

Answer:

a) q = -9.23 cm, b) h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

1 / f = 0.6 / 4 = 0.15

f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

1 / q = 1 / f - 1 / p

1 / q = 1 / 6.67 - 1/24

1 / q = 0.15 - 0.04167 = 0.10833

q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

M = h ’/ h = - q / p

h’= - q / p h

h’= - (-9.23) / 24.0 0.150

h’= 0.05759 cm

h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0

3 years ago

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$

= $1200 \ units$

The total number of particles will be:

= $50+100$

= $150$

hence,

Energy of each A particle or each B particle will be:

= $\frac{1200}{150}$

= $8 \ units$

5 0

3 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

3 years ago

The diagram shows the scales used for recording

padilas [110]

Answer: 212

Explanation:

3 0

3 years ago