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3 years ago

Suppose you increase your walking speed from 5m/s to 15m/s in a period of 1 s. What is your acceleration?

1 answer:

7 0

Acceleration=change in velocity divided by change in time or $a=\frac{\Delta v}{\Delta t}$

we are given that it goes from 5m/s to 15m/s
$\Delta v=v_f-v_0$
in this case, $\Delta v=15\frac{m}{s}-5\frac{m}{s}=10\frac{m}{s}$

and the change in time or $\Delta t$ is 1 second because this change occurs over a period of 1 second

so the acceleration is $a=\frac{10\frac{m}{s}}{1s}=10\frac{m}{s^2}$
the acceleration is $10\frac{m}{s^2}$

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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 4.9 m/s.

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

Jess rides her motorcycle at an average speed of 20 m/s for 500 seconds. how far did she ride

Sedaia [141]

Distance = (speed) x (time)

Distance = (20 m/s) x (500 s)

Distance = (20 x 500) (m·s / s)

Distance = 10,000 m

5 0

3 years ago

6. The image to the right shows a moment of inertia

Trava [24]

The moment of inertia is the rotational analog of mass, and it is given by

the product of mass and the square of the distance from the axis.

The moment of inertia changes as the position of the weight is changed, which indicates that; statement is incorrect

Reasons:

The weight on each arm that have adjustable positions can be considered as point masses.

The moment of inertia of a point mass is I = m·r²

Where;

m = The mass of the weight

r = The distance (position) from the center to which the weight is adjusted

Therefore;

The moment of inertia, I ∝ r²

Which gives;

Doubling the distance from the center of rotation, increases the moment of inertia by factor of 4.

Therefore, the statement contradicts the relationship between the radius of rotation and moment of inertia.

Learn more about moment of inertia here:

brainly.com/question/4454769

7 0

2 years ago

En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)

Virty [35]

Answer:

I only speak English

Explanation:

I'm sorry can you type it in English

7 0

3 years ago

Many industries are powered via distant power stations. Calculate the current flowing through a 7,300m long 10. copper power lin

Oliga [24]

Answer:

Current, I = 1000 A

Explanation:

It is given that,

Length of the copper wire, l = 7300 m

Resistance of copper line, R = 10 ohms

Magnetic field, B = 0.1 T

$\mu_o=4\pi \times 10^{-7}\ T-m/A$

Resistivity, $\rho=1.72\times 10^{-8}\ \Omega-m$

We need to find the current flowing the copper wire. Firstly, we need to find the radius of he power line using physical dimensions as :

$R=\rho \dfrac{l}{A}$

$R=\rho \dfrac{l}{\pi r^2}$

$r=\sqrt{\dfrac{\rho l}{R\pi}}$

$r=\sqrt{\dfrac{1.72\times 10^{-8}\times 7300}{10\pi}}$

r = 0.00199 m

$r=1.99\times 10^{-3}\ m=2\times 10^{-3}\ m$

The magnetic field on a current carrying wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$I=\dfrac{2\pi rB}{\mu_o}$

$I=\dfrac{2\pi \times 0.1\times 2\times 10^{-3}}{4\pi \times 10^{-7}}$

I = 1000 A

So, the current of 1000 A is flowing through the copper wire. Hence, this is the required solution.

4 0

3 years ago