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3 years ago

Plz help, and show work

1 answer:

7 0

1.
Given:
S=-35m(downward), a=-9.8m/s2, u=0(start from rest)

Solution:
Formula: S=ut+(1/2)(a)(t^2)
-35m=(0)t+(1/2)(-9.8m/s)(t^2)
t^2= 7.142857
t= 2.67s

Answer: t=2.67s

2.
Given:
u=0(start from rest), t=20s, v=2m/s

Solution:
Formula: v=u+at
2m/s=0+a(20s)
a=0.1m/s^2

Answer: a=0.1m/s^2

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Light enters a substance from air at 30 degrees to thenormal.

Gwar [14]

Answer:

Explanation:

The angle of incidence and refraction are both measured from the normal

angle of incidence = 30°

angle of refraction = 23°

refractive index(n) = sini / sinr

n = sin30°/sin23°

n = 1.27965

refractive index (n) = 1/sinC

where C is the critical angle.

sinC= 1/n

C =arcsin (1/n)

C =arcsin (1/1.27965)

C = 51.39°

5 0

4 years ago

a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat

mash [69]

The final temperature of the block is $27.5^{\circ} C$

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where:

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

$Q = 2,130 J$ is the amount of energy supplied to the block

$C_s = 710 J/kg^{\circ}C$ is the specific heat capacity of the block

Solving for $\Delta T$ , we find the increase in temperature:

$\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C$

And since the initial temperature was

$T_i = 25^{\circ}C$

The final temperature will be

$T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0

3 years ago

How much work is done if I use 5N of force to move an object 2 meters

sweet-ann [11.9K]

It’s 10 joules. W=FD, W=5•2=10

8 0

3 years ago

Suppose that the loudspeaker in the problem had a mass of 500 kg and the ropes hung 20∘ from the vertical. Into which of the fol

iogann1982 [59]

The following intervals would you expect T(tension) to fall : <u>2000 to 4000 N </u>

<h3>Further explanation</h3>

The force acting on a system with static equilibrium is 0

$\large{\boxed{\bold{\sum F=0}}$

(forces acting as translational motion only, not including rotational forces)

$\displaystyle \sum F_x = 0\\\\ \sum F_y = 0$

We complete the questions:

A 20-kg loudspeaker is suspended 2.0 m below the ceiling by two ropes that are each 30? from vertical.

Find the value of T, the magnitude of the tension in either of the ropes.

Express your answer in newtons.

Suppose that the loudspeaker in the problem has a mass of 500 kg and the ropes hung 20? from the vertical. Into which of the following intervals would you expect T to fall? You don't have to calculate Texactly to answer this question; just make an estimate.

500 to 1000 N

1000 to 2000 N

2000 to 4000 N

4000 to 8000 N

8000 to 16,000 N

In a 500-kg loudspeaker system and two ropes that are each 20° from vertical, we see the forces acting on the y axis (vertical)

$\displaystyle \sum F_y = 0\\\\T1_y+T2_y-w=0\\\\T1~cos~20^o+T2~cos~20^o=m\times g$