How much force is needed to accelerate a 3kg book in 15m/s

Kim throws a beach ball up in the air. It reaches its maximum height 0.50s later. We can ignore air resistance. What was the bea

notka56 [123]

Answer:

The beach ball's velocity at the moment it was tossed into the air is <u>4.9 m/s.</u>

Explanation:

Given:

Time taken by the ball to reach maximum height is, $t=0.50\ s$

We know that, velocity of an object at the highest point is always zero. So, final velocity of the ball is, $v=0\ m/s$

Also, acceleration acting on the ball is always due to gravity. So, acceleration of the ball is, $a=g=-9.8\ m/s^2$

The negative sign is used as acceleration is a vector and it acts in the downward direction.

Now, we have the equation of motion relating initial velocity, final velocity, acceleration and time given as:

$v=u+at$

Where, 'u' is the initial velocity.

Plug in the given values and solve for 'u'. This gives,

$0=u-9.8(0.5)\\u=9.8\times 0.5\\u=4.9\ m/s$

Therefore, the beach ball's velocity at the moment it was tossed into the air is 4.9 m/s

3 0

3 years ago

Read 2 more answers

P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar

creativ13 [48]

Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

3 0

3 years ago

What is a vernier caliper used for?

iris [78.8K]

me ajudem por favor pra agora de noite

5 0

3 years ago

A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th

Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area $A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2$

Let the tensile modulus of Nickel $E = 170 \times 10^9Pa$ .

The elongation of the rod can be calculated using the following formula:

$\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m$

6 0

3 years ago

the winner of a 10km road race took half an hour to complete the race. Calculate the average speed. Give your answer in metres p

leonid [27]

Hello there!

For this:

1). Convert 10km to meters!
2). Convert the 30 minutes into seconds!

3). Use the following formula to solve for speed!

speed= distance/time

Note: The units should automatically work out to m/s. :)

My goal is to make sure you understand the problem, which is why I won't be giving you the answer. It'll be more work now, but less work in the future! :)

Hope this helped!

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4 0

3 years ago