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3 years ago

Discuss the relationship between acceleration and velocity. Are these scalar or vector quantities?

1 answer:

5 0

Acceleration is just speeding up, and velocity is speeding up in a certain direction. If you are changing your velocity you accelerating, as you are changing direction..

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If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit

Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) = (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 = (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0

3 years ago

A spring with k = 19.5 N/cm is initially stretched 1.39 cm from its equilibrium length. a) How much more energy is needed to fur

lara31 [8.8K]

Answer:

Energy needed = 54.02 J

Explanation:

the Energy in an elastic spring from hookes law is given as

F= ke , therefore the energy (E) is

E = $\frac{1}{2} Ke^{2}$

K = 19.5 N/cm

e = 1.39cm

E = $\frac{1}{2}$ x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = $\frac{1}{2}$ x 19.5 x 6.93

E = 67.57 J

The more energy needed for the further stretch is

67.57 - 13.55

Energy needed = 54.02 J

7 0

3 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

garri49 [273]

Answer:

$w_{0}=14$

$t=\frac{\pi }{14}$

Explanation:

Data

mass m= 100g

Length L= 5cm

we can use:

gm-kL= 0

divide both side by m

g - $\frac{kL}{m}$ =0

where

$\frac{k}{m}$ = $\frac{g}{L}$

$\frac{k}{m}=w_{0}$ ^{2}

so now

$w_{0}^{2}$ = $\frac{9.8*100}{5}$

$w_{0}^{2}=\frac{980}{5}$

$w_{0}^{2}=196$

square both side

$w_{0}=\sqrt{196}$

$w_{0}=14$

We can apply:

u(t)=Acoswt +Bsinwt

u(t)=Acos14t +Bsin14t

u(0)=0 where A=0

therefore

u(0) = Bsin14t

$u^{'}$ (0) = 10 ⇒ 10=14B ⇒ B= $\frac{14}{10}$ B= $\frac{5}{7}$

so now u(t)= $\frac{5}{7}$ sin14t

so t will be:

t= $\frac{\pi }{14}$

t= $\frac{3.14}{14}$

t=0.22 seconds

8 0

3 years ago

What process is represented by this redox equation ? 6H 2 O+6CO 2 C 6 H 12 O 6 +6O 2

denis-greek [22]

Answer:

C. Photosynthesis

Explanation:

The redox reaction given in this question is given as follows:

6H2O + 6CO2 → C6H12O6 + 6O2

In this reaction, water (H2O) reacts with carbon dioxide (CO2) to form glucose (C6H12O6) and oxygen (O2). This is the general equation for PHOTOSYNTHESIS process, which is the process whereby plants manufacture their food (sugar) by combining water and carbon dioxide.

In the equation above, water is oxidized into oxygen, while carbon dioxide is reduced into glucose. This makes it an oxidation-reduction (redox) reaction.

6 0

2 years ago

Which type of electronic community allows real-time discussion among members?

Brilliant_brown [7]

Instant messaging is the type of electronic community that allows real-time discussion among members.

Instant messaging (IM) is a type of online chat which offers real-time text transmission over the Internet. A LAN messenger operates in a similar way over a local area network.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

4 0

3 years ago