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3 years ago

An unbalanced electric charge is sometimes called a

1 answer:

7 0

Atoms with unbalanced electrical charges are called ions.

Those with positive charges are called cations.

Those with negative charges are called anions.

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I need help with my physics homework

Strike441 [17]

What do u need help with

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3 years ago

Read 2 more answers

(a) Calculate the height of a cliff if it takes 2.45 s for a rock to hit the ground when thrown straight up from the cliff with

ANEK [815]

An object in free fall motion is under the influence of gravitational force only

The height and time are;

(a) The height of the cliff, h ≈ 8.00 meters

(b) The time it takes the rock to rich the ground with the same speed going downward is approximately 0.67 seconds

The reason the above values are correct are as follows:

The given parameters are;

The time it takes for the rock to hit the ground when thrown straight up from the cliff, t = 2.45 s

The initial velocity with which the rock is thrown, u = 8.75 m/s

(a) To find the height of the cliff

Solution

The kinematic equation of motion is s = y₀ + u·t + (1/2)·g·t²

The time it takes to maximum height, $t_{max}$ = 8.75/9.81

The time it takes to get back to the cliff edge when thrown = 2 × 8.75/9.81

The time taken to fall at the velocity of u downwards from the cliff edge, $t_{cliff}$ , is given as follows;

$t_{cliff}$ = 2.45 - 2 × 8.75/9.81 ≈ 0.666

Height of cliff, h = 8.75×(2.45 - 2 × 8.75/9.81) + (1/2) × 9.81 × (2.45 - 2 × 8.75/9.81)² ≈ 8

The height of the cliff, h ≈ 8.00 meters

(b) The time it takes the rock to rich the ground when thrown straight down art the same speed, (calculates above) = $t_{cliff}$ ≈ 0.67 seconds

Learn more about free fall motion here:

brainly.com/question/13297394

7 0

3 years ago

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest-accelerating animals,

erik [133]

Answer:

Explanation:

given,

speed of cheetah = 27 m/s

time taken = 4 s

mass of cheetah = 110 kg

$a = \dfrac{27-0}{4}$

a = 6.75 m/s²

v² = u² + 2 as

27² = 0 + 2 × 6.75× s

s = 54 m

a) $power = \dfrac{work\ done}{t}$

$power = \dfrac{F.s}{t}$

$power = \dfrac{m.a.s}{t}$

$power = \dfrac{110\times 6.75\times 54}{4}$

$power = 10,023\ watt$

b) $power = \dfrac{10,023}{746}$

$power = 13.46\ horsepower$

5 0

3 years ago

A block with mass m =7.5 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.27 m.

Irina18 [472]

1) The spring constant is 272 N/m

2) The oscillation frequency is 0.96 Hz

3) The speed of the block at t = 0.33 s is 1.7 m/s

4) The maximum acceleration is $25.3 m/s^2$

5) The net force on the block at t = 0.33 s is 173.3 N

6) The potential energy is maximum at the highest point of the oscillation

Explanation:

When the block is hanging on the spring and the spring is in equilibrium, it means that the weight of the block is balancing the restoring force of the spring. Therefore, we can write:

$mg=kx$

where

$mg$ is the weight of the block, with

m = 7.5 kg being the mass

$g=9.8m/s^2$ is the acceleration of gravity

$kx$ is the restoring force in the spring, with

$k$ being the spring constant

x = 0.27 m is the stretching in the spring

Solving for k, we find the spring constant:

$k=\frac{mg}{x}=\frac{(7.5)(9.8)}{0.27}=272 N/m$

The oscillation frequency of a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here we have

k = 272 N/m is the spring constant

m = 7.5 kg is the mass

Substituting, we find the frequency of oscillation:

$f=\frac{1}{2\pi}\sqrt{\frac{272}{7.5}}=0.96 Hz$

The block is pushed downward with speed

$v_0 = 4.2 m/s$

at $t=0$ . This means that we can write the equation of its speed at time t as

$v(t) = -v_0 cos(\omega t)$

where

$\omega = 2\pi f = 2\pi(0.96)=6.02 rad/s$ is the angular frequency of the system

We can verify that for $t=0$ , the equation gives $v(0)=-v_0$ , so it gives the correct initial velocity (the negative sign means downward)

Now we can substitute t = 0.33 s to find the velocity of the block at that time:

$v(0.33)=-(4.2)(cos (6.02\cdot 0.33))=+1.7 m/s$

So the speed is 1.7 s.

The maximum acceleration in a spring-mass system is given by

$a_{max}=\omega^2 A$ (1)

where

$\omega$ is the angular frequency

A is the amplitude

To find the maximum acceleration, we can use also the following relationship:

$v_0 = \omega A$ (2)

Divinding (1) by (2),

$\frac{a_{max}}{v_0}=\omega$

And so we find

$a_{max}= v_0 \omega = (4.2)(6.02)=25.3 m/s^2$

The acceleration of the block at time t is equal to the derivative of the velocity, therefore

$a(t) = \omega^2 A sin(\omega t) = a_{max} sin(\omega t)$

where

$a_{max}=25.3 m/s^2$ is the maximum acceleration of the system

$\omega=6.02 rad/s$ is the angular frequency

By substituting t = 0.33 s, we find the acceleration:

$a(0.33)=(25.3)sin(6.02\cdot 0.33)=23.1 m/s^2$

And therefore, the net force on the block is equal to the product between its mass and its acceleration:

$F=ma=(7.5)(23.1)=173.3 N$

The potential energy of the system is given by the sum of the gravitational potential energy ( $E_g$ ) and the elastic potential energy ( $E_e$ ):

$E=E_g+E_e$

where

$E_g = mgh$ , where

m is the mass of the block

g is the gravitational acceleration

h is the height of the block

$E_e=\frac{1}{2}k(x'-x)^2$ , where

k is the spring constant

(x'-x) is the new stretching of the spring relative to the equilibrium position (x=0.27 m)

We notice that:

$E_e$ is maximum when $(x-x')$ is maximum, so at the highest point of the oscillation and at the lowest point of the oscillation
$E_g$ is maximum when $h$ is maximum, so at the highest point of the oscillation

Therefore, if we add the two contribution, the potential energy is maximum when both $U_g, U_e$ are maximum, so at the highest point of the oscillation.

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3 years ago

Name two technologies that people in developing nations could benefit from.

Verizon [17]

Water filters and internet to get more knowledge

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3 years ago