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klio [65]
3 years ago
6

You just measured a block of wood and obtained the following information: mass = 55.120 g length = 8.5 cm height = 4.3 cm width

= 3.3 cm Determine the volume and density of the wood block.
Physics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

Volume of the wood = 120.615 cm3

Density of the wood = 0.457 g/cm3

Explanation:

By definition, density is the mass of a substance per unit volume and its units are kg/m3, g/l etc.

Volume of a cuboid = length * breadth * height

Length = 8.5 cm

Breadth = 3.3 cm

Height = 4.3 cm

Volume of the wood = 8.5 * 3.3 * 4.3

= 120.615 cm3

Mass of the wood = 55.120 g

Density = mass/volume

= 55.120/120.615

= 0.457 g/cm3

= 457 kg/m3

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The place kicker on a football team kicks a ball from ground level with an initial speed of 5.80 m/s at an angle of 17.0° above
Likurg_2 [28]

Answer:

0.34 s

Explanation:

Given that,

Initial speed of a ball, u = 5.8 m/s

It is kicked at an angle of 17.0° above the horizontal.

The vertical component of velocity will be,

u_y=u\sin\theta\\\\=5.8\times \sin 17\\\\=1.69\ m/s

Let it takes t time in the air before it lands on the ground again. It can calculated as :

t=\dfrac{2u_y}{g}\\\\t=\dfrac{2\times 1.69}{9.8}\\\\t=0.34\ s

So, it will take 0.34 seconds.

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3 years ago
The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

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t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

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a stone is thrown vertically upwards with a velocity of 20 m per second what will be its velocity when it reaches a height of 10
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Here's the info we have:

initial velocity is 20 m/s;

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Answer:

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Explanation:

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3 years ago
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