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Lelu [443]
3 years ago
6

An 0.08 kg arrow is shot from a bow with a velocity of 25 m/s.

Physics
1 answer:
sukhopar [10]3 years ago
5 0
123846284741741741741741852851 to he power of 46
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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

  • 0.829 = <em>x</em> / 10.8
  • <em>y </em>= 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0
3 years ago
Given a 10-V power supply, would a 20-ohm resistor and a 5-ohm resistor need to be arranged in parallel or in series to generate
Bogdan [553]

Answer:

The resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

Explanation:

We are given 10 voltage power source and we have two Resistors with resistance of 20 ohm and 5ohm.

We need to find the orientation in which these two resistors would be arranged so that the circuit could get a current of 2.5Ampere.

Using ohm's law we have

V = I*R

V= voltage

I= current

R= resistance

10 = 2.5*R

R = 10/2.5 = 4ohm

that means we need a total of 4ohm resistance from these two resistors.

since the net Resistance(4ohm) is lower than the smallest resistance(5ohm) available that means the orientation of the resistors will be in parallel.

\frac{1}{R} = \frac{1}{R1} +\frac{1}{R2}\\                 = \frac{1}{20} +\frac{1}{5}\\ \\                 = \frac{1+4}{20} =\frac{1}{4}

R(net) =4ohm

Now the orientation of the resistors are in parallel so the current will be divided.

we know that the current will divide in opposite manner the arm which provides more resistance less current will flow from there and vice versa.

We know that the voltage in parallel remains same

In 20 ohm resistance

again using ohms law

V = i1*R1

10 = i1*20

i1 = 0.5A

in 5ohm resistor

V=i2*R2

10 = I2*5

i2 =2A

and i1+i2 = 0.5+2= 2.5A which means our calculation is correct.

Therefore the resistors will be in parallel to produce a net resistance of 4ohm and current in 20 ohm resistor will be 0.5A and 5ohm resistor will be 2A.

6 0
2 years ago
A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
dmitriy555 [2]

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

F = m * g = d * 4/3π* Rext³ * g

replacing values

F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N

Note:

when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

4 0
3 years ago
Compare the two numbers using &lt;, &gt;, =
Effectus [21]
Vas happenin!!

1. =
2. >
3. <
4. <
5. <
6. >

Hope this helps *smiles*

3 0
2 years ago
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