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marysya [2.9K]
3 years ago
5

-2 5/8 is bigger than -5/12

Physics
1 answer:
Neko [114]3 years ago
8 0

Answer:

hey...................................

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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
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
How do boron-10 and boron-11 differ?
tensa zangetsu [6.8K]

Answer:

I think it has to do something with their ionizations... not entirely sure though.

Explanation:

3 0
3 years ago
I did questions 2 and 3 but I don't know if they are right. Someone help!
Tems11 [23]

You got it right my friend

8 0
3 years ago
Read 2 more answers
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
2 years ago
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