sinat
Accelerationa
2 2.84
7 34
TABLE
in Elination
t2 t3 T2 2
1=0.04
2.29 1.25 1.28 1.271.61
2 460 = 0.00 4.59 1.16 1.081.12 1-25
3 so = 0.12 6.89 0.88
097 0.53
4. = 0.16 9.210.8
9.21 0.850.796. 82/0.67
(So
0.72 0.77 0.75 l 0-56/
0.28
49. al
27. 49
13 11%.
41. 2L
= 0.2 11.54​

Answers

Answer 1

Answer:

so you have a question

Explanation:

either way,you have a nice day


Related Questions

Philosophy: The Big Picture Unit 8

Would an existentialist argue that the study of philosophy was a good use of a life?

A. Yes, if society valued the results of the study.
B. Yes, but only if the individual found it meaningful.
C. No, a life should be spent minimizing the role of anxiety.
D. No, there is no way that philosophy could create a meaningful life.

Answers

the answer to this would be C

Before we make measurements, let's make sure we understand the circuit. 1. Select all of the following that correctly describe what a volt meter and ammeter measure. Select all that apply: A volt meter measures the potential difference (or voltage) across a circuit element. A volt meter measures the potential difference (or voltage) passing through a circuit element. A ammeter measures the electric current passing through a circuit element. A ammeter measures the electric current across a circuit element.

Answers

Answer:

the correct answers are a and c

Explanation:

In an electrical circuit there are two important quantities to measure, such as voltage and current.

Voltage is the potential difference between two points in a circuit

current is the number of electrons you pass through a given point per unit of time.

Now let's analyze each answer

a) true. The potential difference across an element

b) False. The potential difference is u field there is no physical entity that moves

c) True. The current is electrons in motion and these pass through the given element

d) False. There is a physical quantity that passes through the point

the correct answers are a and c

A 5kg block rests on a 30° incline. The coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding. a) up the incline, b) down the incline ? ​

Answers

Answer:

Hope It Help

Explanation:

That's all I know

Why do astronomers use frequencies other than the visible ones when they are
investigating the universe?

Answers

Because not all of the universe can be seen with a visible spectrum

1 kg block slides down a frictionless inclined plane that makes an angle of 300 with respect to the ground. The total length of the plane is 2 m, but midway down it collides with a second block, weighing 0.5 kg. The two blocks stick together and travel as one unit the rest of the way down the ramp. What is the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane

Answers

Answer:

the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

Explanation:

Given that the data in the question;

angle of inclination with respect to the ground [tex]\theta[/tex] = 30°

length of plane d = 2m

m₁ = 1 kg

m₂ = 0.5 kg

now, velocity of the first block at midpoint;

[tex]\frac{1}{2}[/tex]mv² = mgsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

[tex]\frac{1}{2}[/tex]v² = gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex]

v² = gsin[tex]\theta[/tex]d

v = √( gsin[tex]\theta[/tex]d)

g is 9.8 m/s

so we substitute

v = √( 9.8 × sin30° × 2)

v = √( 19.6 )

v =  3.13 m/s

Now, velocity just after collision of the blocks will be;

(m₁ + m₂)v₂ = m₁v

v₂ = m₁v / (m₁ + m₂)

we substitute

v₂ = (1 × 3.13) / (1 + 0.5)

v₂ = 3.13 / 1.5

v₂ = 2.0866 m/s

now, final kinetic energy will be;

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + Initial Kinetic energy

[tex]KE_f[/tex] = (m₁ + m₂)gsin[tex]\theta[/tex][tex]\frac{d}{2}[/tex] + [tex]\frac{1}{2}[/tex]mv₂²

we substitute

[tex]KE_f[/tex] = [(1 + 0.5)9.8 × sin30 × [tex]\frac{2}{2}[/tex]] + [[tex]\frac{1}{2}[/tex] × 1.5 × 2.0866 ]

[tex]KE_f[/tex] = 7.35 + 3.2654

[tex]KE_f[/tex] = 10.62 J

Therefore, the kinetic energy of the combined 1.5 kg block when it reaches the bottom of the plane is 10.62 J

A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This force causes the house to move from rest to an upward speed of 15 cm/s in 5.0 s. What is the mass of the house?​

Answers

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

[tex]F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg[/tex]

So, the mass of the house is equal to 95000 kg.

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