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never [62]
2 years ago
9

I need this ASAP will give brainliest Question 2

Physics
1 answer:
myrzilka [38]2 years ago
4 0

Answer:

2a. 24 m³

2b. 19200 Kg

2c. 192000 N

2d. 16000 N/m²

Explanation:

2a. Determination of the volume of paraffin.

Length (L) = 4 m

Width (W) = 3 m

Height (H) = 2 m

Volume (V) =?

V = L × W × H

V = 4 × 3 × 2

V = 24 m³

Therefore the volume of paraffin is 24 m³

2b. Determination of the mass of paraffin.

Volume (V) = 24 m³

Density (ρ) = 800 Kg/m³

Mass (m) =?

ρ = m / V

800 = m / 24

Cross multiply

m = 800 × 24

m = 19200 Kg

Therefore, the mass of paraffin is 19200 Kg

2c. Determination of the weight of paraffin.

Mass (m) = 19200 Kg

Acceleration due to gravity (g) = 10 N/Kg

Weight (W) =?

W = m × g

W = 19200 × 10

W = 192000 N

Therefore, the weight of paraffin is 192000 N

2d. Determination of the pressure.

Density (ρ) = 800 Kg/m³

Depth (h) = 2 m

Acceleration due to gravity (g) = 10 N/Kg

Pressure (P) =?

P = ρgh

P = 800 × 10 × 2

P = 16000 N/m²

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In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

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2 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
slega [8]

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

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             F = 7.37 10⁵ N

.d) the ratio of this force to weight

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             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

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