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alina1380 [7]
4 years ago
15

For a constant force, the impulse could be increased by: A.decreasing the distance. B.keeping the time the force acts constant.

C.increasing the time the force acts. D.none of the other answers.
Physics
1 answer:
Andre45 [30]4 years ago
5 0
C. Increasing the time of the force
This is because
Impulse = FΔt
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Ivan walks 10 meters west to the water fountain, then runs 2 meters east to his class to avoid a lockout. His displacement would
otez555 [7]

Answer:

<u>-8</u>

Explanation:

if he starts at ten and takes 10 steps left he'll be at -10... then if he takes 2 steps to the right , he's at -8 on the number-line

7 0
3 years ago
Read 2 more answers
If a cylindrical space station 275 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) mus
sergij07 [2.7K]

The centripetal acceleration is responsible for the artificial gravity because the acceleration of an object moving in constant circular motion causing from net external force is called centripetal acceleration. It defines to the center or seeking the center.

Given the following:

Cylindrical space station diameter           = 275 meters; 137.5 meters for the radius

Standard gravity                                       = 9.80665 m/s²

 

Using the formula:

w² x r =g

w² = g / r

w² = 9.80665 m/s² / 137.5 m

w² = 9.80665 m/s² / 137.5 m

w² = 0.0713 s²

Then take the roots

w = 0.267 this is radians per second / 2 x (3.1416 which is the pi)

w = 0.0424 rps convert to rpm

w = 0.0424 r/s (1minute / 60 seconds)

w = 7.08 x 10⁻⁴ revolutions per minute

4 0
3 years ago
A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
ValentinkaMS [17]

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

4 0
3 years ago
A vehicle travelling at an initial velocity of 20km/hr,accelerates at 4m/s².calculate its final velocity after 10 seconds.​
Nikitich [7]

acceleration = Velocity changes ÷ time of the velocity changes

4 m/s^2 =

4 × 10^(-3) × 3600 km / h =

4 × 3.6 =

14.4 km / h

Thus :

14.4 = V(2) - V(1) / t(2) - t(1)

14.4 = V(2) - 20 / 10

Multiply both sides by 10

10 × 14.4 = 10 × ( V(2) - 20 ) / 10

144 = V(2) - 20

Add both sides 20

144 + 20 = V(2) - 20 + 20

V(2) = 164 Km/h

Thus the final velocity after 10 seconds is 164 Km/h .

4 0
3 years ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
3 years ago
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