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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

2 answers:

dalvyx [7]3 years ago

4 0

Answer:

The answer is A

Explanation:

I just C in and it was wrong, The answer is A.

Nonamiya [84]3 years ago

3 0

Based on the graph, the soccer ball is not moving from C) from 4.5 to 5.0 seconds

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<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

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This problem is about Average Speed

We will use this following formula:

$\texttt{Average Speed} = \texttt{Distance} \div \texttt{Time Taken}$

From 4.5 to 5.0 seconds , there is no change in distance of the ball.

t₁ = 4.5 → s₁ = 65 m

t₂ = 5.0 → s₂ = 65 m

$\texttt{Average Speed} = (65 - 65) \div (5.0 - 4.5)$

$\texttt{Average Speed} = 0 \div 0.5$

$\texttt{Average Speed} = 0 \texttt{ m/s}$

Since the average speed is zero, the ball doesn't move from 4.5 to 5.0 seconds.

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<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

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Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

#LearnWithBrainly

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The full question is:
On a keyboard, you strike middle C, whose frequency is 256 Hz. What is the period of one vibration of this tone?
The period of a vibration is the time it takes for the particle to make one full oscillation. Frequency is by definition number of full oscillations per unit of time.
When the frequency is expressed in Hz that unit of time is one second.
So there is the following relation between frequency and period:
$T=\frac{1}{f}$
When we plug in the numbers we get:
$T=\frac{1}{256}=0.0039$s$

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3 years ago

A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

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3 years ago

A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight

photoshop1234 [79]

Answer:

$t=1.9 sec$

Explanation:

From the question we are told that:

Height $h=28m$

Time $t=3s$

Generally the Newton's equation for Initial velocity upward is mathematically given by

$s=ut+\frtac{1}{2}at^2$

$28=3u-\frac{1}{2}*9.8*3^2$

$u=24.03m/s$

Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28

$v=\sqrt{24.03-2*9.8*28}$

$v=5.35m/s and -5.35m/s$

Generally the Newton's equation for time to reach initial velocity is mathematically given by

$v=u+at$

$5.35=24.03-9.8t$

$t=\frac{28.03-5.35}{9.8}$

$t=1.9 sec$

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2 years ago

A sled is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy isJ. (Formula: PE =

elixir [45]

Data:

h = 2m
m = 45 Kg
PE = ? (Joule)

Adopting, gravity (g) ≈ 9,8 m/s²

Formula: $PE_{grav} = mass * g * height$

Solving:

$PE_{grav} = mass * g * height$
$PE_{grav} = 45*9,8*2$
$PE_{grav} = 882J$

Answer:
The sled's potential energy is 882 Joules

8 0

2 years ago

Read 2 more answers

If javier has a car thats 250,000 kg and a forse of 300 newtons.what would be the accleration.

const2013 [10]

Answer:

The acceleration would be 3.455.

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3 years ago