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yawa3891 [41]
3 years ago
8

Date

Physics
1 answer:
shtirl [24]3 years ago
4 0

Answer:

W=F×H,Power=W/Time

Explanation:

W=Force ×Height

750N×16 =12000

Power=Work/Time

12000/5=2400J

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<h2><u>Answer:</u></h2><h2>A. Displacement</h2><h2>B. Time</h2><h2 /><h2>Your Welcome ❤</h2>
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3 years ago
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Which cycle causes earthquake and volcanic activity? carbon water tectonic phosphorus
erik [133]
The answer would be tectonic. hope this helps (:
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3 years ago
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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

5 0
3 years ago
A force, 10 N drags a mass 10 kg on a horizontal table with an acceleration of 0.2
kifflom [539]

Answer:

20&£+)##&843&()-_££-()&_2+0&&-£_!)

4 0
3 years ago
An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above
ryzh [129]

Answer:

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v  that is

v² = u² - 2gh      ............1        for upward speed

v² = u² + 2gh     ............2        for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9  = 26.76

v² = 8² + 2(9.8) 1.9   = 101.24

v = 5.173  m/s    ..............3

v = 10.061 m/s   ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173  = 2.827 m/s     .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s     ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

6 0
3 years ago
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