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marshall27 [118]
3 years ago
9

Solve A and B using energy considerations.

Physics
1 answer:
Alisiya [41]3 years ago
7 0
 <span>Use the kinematic equation vf^2 = vi^2 + 2ad where; 
vf = ? 
vi = 0 m/s 
a = 9.8 m/s^2 
d1 = 10 m 
d2 = 25 m 

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s 
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s 
</span>
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If he’s walking at a constant velocity there is no acceleration.
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3 years ago
A diver is on a board 1.80 m above
Virty [35]

Answer:

v = 6.95 m/s

Explanation:

Given that,

A diver is on a board 1.80 m above  the water, s = 1.8 m

The initial speed of the diver, u = 3.62 m/s

Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :

v^2-u^2=2gs\\\\v=\sqrt{u^2+2gs} \\\\v=\sqrt{(3.62)^2+2(9.8)(1.8)} \\\\v=6.95\ m/s

So, she will hit the water with a speed of 6.95 m/s.

8 0
3 years ago
Read 2 more answers
An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use
shtirl [24]

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object  = 200 feet/second

Final velocity of object  = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

a=\frac{v_f-v_i}{t}

where vf represents final velocity, vi represents initial velocity and  is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a= -30m/s

The acceleration of the object is -30m/s  

3 0
3 years ago
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and
gavmur [86]

Answer:

Density of Sand is 2.653g/cm^{3}.

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle(w_{max})=48.4gm-23.5gm

w_{max}=24.9g

Since we know density of water d_{w} =1g/cm^{3} and w_{max}

We can calculate volume of empty space in the bottle(V).

w_{max}=d_{w}\timesV

V=\frac{w_{max} }{d_{w} }

V=\frac{24.9}{1}

V=24.9 g/cm^{3}

Now bottle is partially filled with sand,and weight of bottle is (w_{s})36.5gm

So,

Amount of sand added (m_{s})=36.5-Weight of the bottle

m_{s}=13g

After filling the bottle with water again,the weight of the bottle becomes (W_{2}=56.5g)

Therefore,

amount of water added to the bottle of sand in grams = W_{2}-36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/cm^{3}

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle(V_{w})=20cm^{3}

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V-V_{w}

V_{s}=24.9g-20g

V_{s}=4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = \frac{m_{s} }{V_{s} }

density of sand =\frac{13}{4.9}

density of sand = 2.653g/cm^{3}

8 0
3 years ago
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