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zhenek [66]
3 years ago
9

I need to know number 10

Physics
1 answer:
Mkey [24]3 years ago
8 0
Displacement was 0, Alex went north 1200 then south 1200 and east 600 then west 600, leaving himself at his starting point. He covered 1200+1200+600+600 in distance, or 3600 meters (or 3.6km when converted)
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In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a
IrinaK [193]
A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2


B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

= 224,175N
7 0
3 years ago
What is the power output of a pump which can raise 60kg of water height to height of 10m every minute
sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0
2 years ago
A wire that is 0.65 m long and carrying a current of 8.2 A is at right angles to a uniform magnetic field. The force on the wire
Luda [366]

Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

F=ILB sin \theta

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

\theta is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

\theta=90^{\circ} since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T

3 0
3 years ago
A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for
Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

8 0
3 years ago
An 80.0 kg skier slides down a hill shaped as shown. Assume
umka21 [38]

The height above the ground from where the skier start is 11.5 m.

<h3>Conservation of energy</h3>

The height above the ground from where the skier start is determined by applying the principle of conservation of energy as shown below;

P.E = K.E

mgh = ¹/₂mv²

gh = ¹/₂v²

h = \frac{v^2}{2g} \\\\h = \frac{15^2}{2 \times 9.8} \\\\h = 11.5 \ m

Thus, the height above the ground from where the skier start is 11.5 m.

Learn more about conservation of energy here: brainly.com/question/166559

8 0
3 years ago
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