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3 years ago

Plz I really need this question!!!!

1 answer:

7 0

I have a hunch that you've got the question totally covered, and what you actually need is the answer.

Kinetic energy = (1/2) (mass) (speed²)

Multiply each side by 2 : 2 x KE = (mass) ( speed²)

Divide each side by (mass): Speed² = 2 x KE / mass

Square root each side: Speed = √(2 KE/mass)

Look at that ! The question GIVES you the KE and the mass. All you have to do is plug those 2 numbers into the right side of that equation, turn the crank, do the arithmetic, and the speed falls out.

I get 200 m/s . You need to check my work.

(IF that's correct or anywhere close, it's equivalent to something around 447 miles per hour, which is very reasonable for a cruising airliner.)

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Which term describes the image of an object that is placed in front o a convex mirror?

Dmitriy789 [7]

Answer:

The answer of this question is :- Virtual image

5 0

3 years ago

Read 2 more answers

What mass of steam at 100°C must be mixed with 119 g of ice at its melting point, in a thermally insulated container, to produce

N76 [4]

Answer: $M=27.92\ gm$

Explanation:

Given

mass of ice $m=119\ gm$

Final temperature of liquid $T_f=57^{\circ}C$

Specific heat of water $c=4186\ J/kg-K$

Latent heat of fusion $L=333\ kJ/kg$

Latent heat of vaporization $L_v=2256\ kJ/kg$

Suppose M is the mass of steam at $100^{\circ} C$

Heat required to melt ice and convert it to water at $57^{\circ}C$

$Q_1=mL+mc(T_f-0)$

Heat released by steam

$Q_2=ML_v+Mc(100-T_f)$

$Q_1$ and $Q_2$ must be equal as the heat gained by ice is equal to Heat released by steam

$Q_1=Q_2$

$\Rightarrow mL+mc(T_f-0)= ML_v+Mc(100-T_f)$

$\Rightarrow M=\dfrac{m[L+c\times T_f]}{L_v+c(100-T_f)}$

$\Rightarrow M=\dfrac{119[333\times 10^3+4186\times 57]}{2256\times 10^3+4186\times (100-57)}$

$\Rightarrow M=119\times 0.2346$

$M=27.92\ gm$

7 0

3 years ago

The figure shows the arrangement of master cylinder and slave cylinder of a part of braking system. The master cylinder piston,

Neko [114]

Answer:

a) 35 kPa

d) 140 N

c) 1) Increasing the brake fluid pressure

2) Increasing the slave piston surface area.

Explanation:

The parameters given are;

a) Force applied to the master cylinder piston = 28 N

Cross sectional area of the master cylinder piston = 8 cm² = 8 × 10⁻⁴ m²

The pressure P on the brake fluid is given by the formula for pressure as follows;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{28 \, N}{8 \times 10^{-4} \, m^2} = 35,000 \, N/m^2 = 35,000 \, Pa$

The pressure on the brake fluid, P, produced by the master cylinder piston = 35,000 Pa = 35 kPa

b) Given that the area of the slave piston = 40 cm² = 0.004 m², we have from the formula for pressure, P;

$P= \dfrac{Applied \ force}{Area \over \ which \ force \ is \ applied} = \dfrac{Applied \ force}{4 \times 10^{-3} \, m^2} = 35,000 \, N/m^2$

Therefore;

Applied force on the slave piston = 4 × 10⁻³ m² × 35,000 N/m² = 140 N

c) The force, F produced by the slave cylinder piston is given by the relation;

F = Pressure × Area

Therefore, the two ways of increasing the force produced by the slave cylinder piston is as follows;

1) Increasing the pressure in the brake fluid by increasing the force exerted by the master cylinder piston

2) Increasing the surface area of the slave piston.

4 0

3 years ago

How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from th

-Dominant- [34]

Answer:

W = 7591.56 J

Explanation:

given,

distance of the box, d = 37 m

Force for pulling the box, F = 217 N

angle of inclination with horizontal,θ = 19°

We know,

Work done is equal to product of force and the displacement.

W = F.d cos θ

W = 217 x 37 x cos 19°

W = 7591.56 J

Hence, the work done to pull the box is equal to W = 7591.56 J

8 0

3 years ago

A particular spring has a spring constant of 25 N/m.

kotegsom [21]

Answer:

$e.p.e = \frac{1}{2} k {x}^{2} \\ = \frac{1}{2} \times 25 \times (0.015m)^{2} \\ = 0.0028125$

0.0028125N

8 0

3 years ago