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A person can run 400m in 72 s.. At what speed are they running

rewona [7]

Speed = distance/time
= 400/72
=5.55 m/s

6 0

3 years ago

a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte

IgorLugansk [536]

From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.

dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

$1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.

7 0

3 years ago

g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell

olga55 [171]

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

The current density is $J = 4*10^6 \ A/m^2$

The conductivity is $\sigma = 2*10^{7} \ S/m$

Generally conductivity is mathematically represented as

$\sigma = \frac{l}{RA}$

Here R is the resistance which is mathematically represented as

$R = \frac{V}{I}$

Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

=> $\sigma = \frac{l}{\frac{V}{J}}$

=> $V = \frac{l * J}{\sigma }$

=> $V = \frac{10 * 4*10^6}{2*10^{7} }$

=> $V = 2 V$

5 0

3 years ago

How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer?

antiseptic1488 [7]

Answer:

1. In general, the best shields will be able to block a spectrum of radiation. Aboard the space station, the use of hydrogen-rich shielding such as polyethylene in the most frequently occupied locations, such as the sleeping quarters and the galley, has reduced the crew's exposure to space radiation.

2. It absorbs harmful ultraviolet radiation from the sun, helps keep Earth's surface warm via the greenhouse effect, and reduces temperature extremes between day and night. ... So, thanks to gravity, although some of Earth's atmosphere is escaping to space, most is staying here.

hope it helps. please mark me as brainliest and follow me ❤️

3 0

3 years ago

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sph

Olenka [21]

Answer:

B.The charge on A is -q; there is no charge on B.

Explanation:

We are given that

Charge=+q

We have to find the correct statement.

When positive charge is placed at center of uncharged metal sphere

insulated from the ground then negative charge(-q) induced on inner

surface A of sphere and the outer surface B is grounded then the surface is neutral .

It means there is no charge on surface B.

Hence, option B is true .

B.The charge on A is -q; there is no charge on B.

4 0

3 years ago