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3 years ago

DESPERATEA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Physics

1 answer:

77julia77 [94]3 years ago

5 0

Answer:

i=1250000000Nm

ii=50000m/s

Explanation:

energy=force ×distance

500000N×(2.5×1000)m

=1250000000Nm

ii. Force=mass ×acceleration

500000N =40000kg ×a

500000 ÷40000=12.5a

s=1/2at^2

2500m=1/2×12.5×t^2

2500m=6.25×t^s

2500÷6.25=400

t^2=400

√t^2=√400

t=20seconds

speed=distance ×time taken

=2500m × 20s

=50000m/s

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How much energy, in Joules, is needed to raise the temperature from 25.87°C to 33.16°C in a 2.66 kg block of each of the followi

RUDIKE [14]

Answer:

(a) Silver, H = 455.67 J

(b) Iron, H = 870.67 J

Explanation:

mass of block, m = 2.66 kg = 266 g

T1 = 25.87° C

T2 = 33.16° C

ΔT = T2 - T1 = 33.16 - 25.87 = 7.29° C

(a) Specific heat of silver, c = 0.235 J/g K

The heat required is given by

H = m x c x ΔT

H = 266 x 0.235 x 7.29 = 455.67 J

(b) Specific heat of iron, c = 0.449 J/g K

The heat required is given by

H = m x c x ΔT

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3 years ago

We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with

amm1812

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

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3 years ago

The element carbon has 3 naturally occurring isotopes. About 99% of carbon isotopes are C-12, about 1% are C-13, and a tiny amou

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Average atomic mass of carbon is 12.01 amu

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4 years ago

A golf club rotates 215 degrees and has a length (radius) equal to 29 inches. The time it took to swing the club was 0.8 seconds

vichka [17]

Answer:

The average linear velocity (inches/second) of the golf club is 136.01 inches/second

Explanation:

Given;

length of the club, L = 29 inches

rotation angle, θ = 215⁰

time of motion, t = 0.8 s

The angular speed of the club is calculated as follows;

$\omega = (\frac{\theta}{360} \times 2\pi, \ rad) \times \frac{1}{t} \\\\\omega = (\frac{215}{360} \times 2\pi, \ rad) \times \frac{1}{0.8 \ s} \\\\\omega = 4.69 \ rad/s$

The average linear velocity (inches/second) of the golf club is calculated as;

v = ωr

v = 4.69 rad/s x 29 inches

v = 136.01 inches/second

Therefore, the average linear velocity (inches/second) of the golf club is 136.01 inches/second

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3 years ago