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Ede4ka [16]
3 years ago
11

PSYCHOLOGY! A _________ is a graphical representation of association between variables.

Physics
2 answers:
pshichka [43]3 years ago
5 0

Answer:

A. Scatterplot

Explanation:

because it is

vichka [17]3 years ago
4 0

Answer:

A. scatterplot.

Explanation:

A scatterplot is a graphical representation of association between variables.

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Four vehicles approach an intersection with a 4 way stop at the same time. Car B is ahead of Car A and both are in the same lane
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Answer: car D

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3 years ago
How is grounding a positive object different from grounding a negative object?
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Answer:

Grounding a Positively Charged Object

Electrons were transferred from the electroscope to the ground. As in the case of grounding a negatively charged electroscope, the grounding of a positively charged electroscope involves charge sharing. The excess positive charge is shared between the electroscope and the ground.

Explanation:

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3 years ago
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3 years ago
Read 2 more answers
The boom of a fire truck raises a fireman (and his equipment – total weight 280 lbf) 60 ft into the air to fight a building fire
Katen [24]

Answer:

a)Work done by fireman=   2.15  Btu

b) Time t= 0.86 sec

Explanation:

Given that

Weight = 280 lbf

We know that 1 lbf = 4.44 N

so 280 lbf = 1245.5 N

Weight =1245.5 N

Height h = 60 ft

We know that

1 ft = 0.3048 m

So 60 ft = 18.28 m

 h =18.28 m

Power = 3.5 hp

We know that

1 hp =0.74 KW

So 3.5 hp = 2.61 KW

Power = 2.61 KJ/s

So the work done by fireman = Weight x h

Now by putting the values

Work done by fireman= 1245.5 x 18.28 J

Work done by fireman=   2267.74 J

Work done by fireman=   2.26774  KJ

We know that 1 Btu= 1.05 KJ

So   2.266 KJ = 2.15 Btu

Work done by fireman=   2.15  Btu

We know that ,rate of work is called power.

Power x time =  work

2.61 x t = 2.26

So t= 0.86 sec

6 0
3 years ago
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
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