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3 years ago

A ball is thrown w a speed of 30m/s at an angle of 10. When is the vertical component of the velocity equal to zero

Physics

2 answers:

insens350 [35]3 years ago

7 0

Answer:

at the top of the ball’s trajectory

Explanation:

irga5000 [103]3 years ago

5 0

Now the vertical velocity of the ball thrown at an angle 10° is given as

Voy(initial vertical velocity)= 30m/s x sin 10

Voy(initial vertical velocity)= 5.2m/s

Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.

Let Vy be the final velocity and that is equal to zero in this case.

Now

Vy= Voy- tx9.8

Where t is the time at which the vertical velocity becomes 0.

Substituting the values we get

0= 5.2-tx9.8

9.8t=5.2

t=0.53 secs

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a 10.0 kg sphere is released from rest in an ocean. as it falls, the water applies a resistive force r

dimaraw [331]

The calculated coefficient of kinetic friction is 0.33125.'

The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.

given mass of the block=10 kg

spring constant k= 2250 Nm

now according to principal of conservation of energy we observe,

the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.

mgh= μ (mgl) +1/2 kx²

10 x 10 x 3= μ(600) +(1125) (0.09)

μ(600) =300 - 101.25

μ = 198.75÷600

μ =0.33125

The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)

Learn more about kinetic friction here-

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4 0

2 years ago

The electric field in a region of space increases from 0 0 to 3450 N/C 3450 N/C in 4.40 s. 4.40 s. What is the magnitude of the

slavikrds [6]

Answer: B = 1380T

Explanation: please find the attached file for the solution

7 0

3 years ago

ou are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As soo

madam [21]

Till the time car is just adjacent to the bicycle we can say

distance moved by cycle = distance moved by car

Time taken by car to accelerate from rest

$t = \frac{v_f - v_i}{a}$

$t = \frac{49 - 0}{7} = 7 s$

Time taken by cycle to accelerate

$t = \frac{23 - 0}{15} = 1.53 s$

now the distance moved by cycle in time "t"

$d = \frac{23 + 0}{2}*1.53 + 23(t - 1.53)$

distance moved by car in same time

$d = \frac{7t + 0}{2}(t)$

now make them equal

$3.5t^2 = 17.595 - 35.19 + 23t$

$3.5 t^2 - 23t + 17.595 = 0$

$t = 5.68 s$

so cycle will move ahead of car for t = 5.68 s

8 0

3 years ago

Read 2 more answers

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago

What is the force required to produce an acceleration of 46.4

Brilliant_brown [7]

That depends on the mass of the object, and the unit of the '46.4' .

If the '46.4' is ' meters per second² ' , then the force required is

(mass of the object in kilograms) x (46.4) newtons .

7 0

3 years ago