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Anestetic [448]
3 years ago
13

A person is standing on a scale in an unmoving elevator. The elevator starts to move upwards at 1 m/s squared. Is the scale read

ing more, less, or the same compared to when the elevator wasnt moving?
Physics
1 answer:
timama [110]3 years ago
7 0

Answer: GREATER

Explanation:when elevator does not move it reads weight of the person . when elevator moves up let apparent weight be F . W acts downwards so net force is F-W

HENCE

F-W =ma

F= ma+W

AS a= 1 m/s^2

F = m (1)+W

HENCE GREATER

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A car is 200 m from a stop sign and traveling toward the sign at 40.0 m/s. At this time, the driver suddenly realizes that she m
victus00 [196]

Answer:

The acceleration of the car will be a=9600m/sec^

Explanation:

We have given that distance from stop sign s = 200 m

Time t = 0.2 sec

We have to find the constant acceleration

Now from second equation of motion s=ut+\frac{1}{2}at^2

200=40\times 0.2+\frac{1}{2}\times a\times 0.2^2

a=9600m/sec^

So the acceleration of the car will be a=9600m/sec^

6 0
3 years ago
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 2
FinnZ [79.3K]

Answer:

Explanation:

The lift is going down with acceleration

Initial speed u = 0

Final speed v = 6 m/s

distance s = 15.25 m

acceleration a = ?

v² = u² + 2 a s

6² = 0 + 2 x a x 15.25

a = 1.18 m /s²

Elevator is going down with acceleration  .

mg - T = ma where T is tension in the cable .

722 x 9.8 - T = 722 x 1.18

7075.6 - T = 851.96

T = 6223.64 N .

7 0
3 years ago
In Youngs double slit experiment the distance between the screen and the slits is 1.00 m. The slits separation is 0.050 mm and t
tester [92]

Check attached photo

8 0
3 years ago
If Noah runs three laps around a 200 meter track in two minutes, what is the average speed of Noah in m/s?
devlian [24]
• 3 laps of 200 meter track = 600 meters
• 600 meters ran in 2 minutes
• 2 minutes = 120 seconds
• 600 meters / 120 seconds = 5 m/s on average
6 0
2 years ago
A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m
Keith_Richards [23]

Answer:

42.69 N and 18.07 N

Explanation:

We are given that

Mass of ladder=6.2 kg

Length of ladder=1.97 m

Distance of Sawhorse A from one end=0.64 m

Distance of sawhorse B from other end=0.17 m

Let center of Ladder=\frac{1.97}{2}=0.985 m

Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m

Distance of sawhorse B from center of ladder=0.985-0.17=0.815  m

Force one ladder due to gravity=mg=6.2\times 9.8=60.76N

Where g=9.8 m/s^2

Torque applied on Sawhorse A=0.345F_a

Torque applied on Sawhorse B=0.815F_b

In equilibrium

0.345F_a=0.815F_b

F_b=\frac{0.345}{0.815}F_a

Total force=F_a+F_b

F_a+\frac{0.345}{0.815}F_a=60.76

\frac{0.815F_a+0.345F_a}{0.815}=60.76

\frac{1.16}{0.815}F_a=60.76

F_a=\frac{60.76\times 0.815}{1.16}=42.69 N

F_b=\frac{0.345}{0.815}\times 42.69=18.07 N

8 0
3 years ago
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