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just olya [345]
3 years ago
12

An upward force of 27.5 n is applied to a string to lift a ball with a mass of 2.5 kg.

Physics
1 answer:
Rudiy273 years ago
4 0
We first find that weight W of the ball
where W= ma
W= (2.5)(9.81)
= 24.525
therefore the net force acting upon the ball is
= upward force - weight
= 27.5 - 24.525
= 2.975 N
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A skateboarder has an acceleration of −1.9 m/s2. If her initial speed is 6 m/s, how long does it take her to stop?
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You said that she's losing 1.9 m/s of her speed every second.

So it'll take

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What are the 3 Newton's law of motions?
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Answer:

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4 0
2 years ago
A place kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. half the crowd hopes the ball will clear
trapecia [35]
<span>The ball clears by 11.79 meters Let's first determine the horizontal and vertical velocities of the ball. h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s Now determine how many seconds it will take for the ball to get to the goal. t = 36.0 m / 15.04 m/s = 2.394 s The height the ball will be at time T is h = vT - 1/2 A T^2 where h = height of ball v = initial vertical velocity T = time A = acceleration due to gravity So plugging into the formula the known values h = vT - 1/2 A T^2 h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2 h = 42.92 m - 4.9 m/s^2 * 5.731 s^2 h = 42.92 m - 28.0819 m h = 14.84 m Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
4 0
3 years ago
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i
kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i  + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

So

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0
3 years ago
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