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Alja [10]
3 years ago
14

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance o

f 2.7 meters. Calculate the vertical distance by which the player will miss the target if he throws the dart horizontally, in line with the dartboard.A.
0.08 meters
B.
0.16 meters
C.
0.32 meters
D.
1.8 meters
Physics
1 answer:
inysia [295]3 years ago
3 0

Answer:

vertical distance, s = 0.16 meters

Explanation:

It is given that,

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard.

Dashboard is placed at a distance of 2.7 meters.

Calculating time from velocity and distance i.e. t=\dfrac{d}{s}=\dfrac{2.7}{15}\ s=0.18\ s

Using second equation of motion for finding vertical distance :

s=ut+0.5gt^2

Here, u = 0 (for vertical velocity )

s=0.5gt^2  

s=0.5\times 10\times (0.18)^2        

s = 0.162 meters

or s = 0.16 meters

Hence, the vertical distance by which the player will miss the target if he throws the dart horizontally, in line with the dartboard is 0.16 meters.

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A 30 ohm resistor and a 20 ohm resistor are
Reptile [31]

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

  • I = V/Rt........... Equation 1

Where:

  • I = Electric current passing through the 30 ohms resistor
  • V = Voltage
  • Rt = Total or effective resistance of the resistors.

From the question,

Given:

  • V = 100 volts
  • Rt = (30+20) ohms (since both resistors are connected in series)
  • Rt = 50 ohms

Substitute these values into equation 1

  • I = 100/50
  • I = 2 A

Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

8 0
3 years ago
The human tibia breaks under an impulse of 55 Ns. If after falling a person typically comes to eat in a time span of 0.005 s how
ozzi

Answer:

0.73 m/s

Explanation:

From Newton second law of motion,

I = m(v-u)...................... Equation 1

Where I = Impulse, m = mass of the person, v = final velocity, u = Initial velocity.

make v the subject of the equation

v =(I/m)+u................. Equation 2

Note: u = 0 m/s as the person is falling from an height.

Given: I = 55 Ns, m = 75 kg, u = 0 m/s

Substitute into equation 2

v = 55/75

v = 0.73 m/s

5 0
4 years ago
An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel
Aleonysh [2.5K]

Answer:

L = 0.635m

Explanation:

This problem involves the concept of stationary waves in pipes. For pipes closed at one end,

The frequency f = nv/4L for n = 1,3,5....n

For pipes open at both ends

f = nv/2L for n = 1,2,3,4...n

Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.

The film solution can be found in the attachment below.

8 0
3 years ago
If you drop a bouncing ball from a height of 40 centimeters, explain why it can only bounce back up to a height of less than 40
iren2701 [21]

Answer:

Due to energy loss while collision ball will not reach to same height while if there is no energy loss then in that case ball will reach to same height

Explanation:

As we know that initially ball is held at height h = 40 cm

So here we can say that kinetic energy of the ball is zero and potential energy is given as

U = mgH

now when strike with the ground then its its fraction of kinetic energy is lost in form of other energies

So the ball will left rebound with smaller energy and hence it will reach to height less than the initial height

While if we assume that there is no energy loss during collision then in that case ball will reach to same height again

8 0
3 years ago
A rifle bullet with mass ma = 8.00 g strikes and embeds itself in a block with mass mb = 0.992 kg that rests on a frictionless,
Doss [256]

Answer:

the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

Explanation:

a)  Kinetic energy of block = potential energy in spring  

½ mv² = ½ kx²

Here m stands for combined mass (block + bullet),

which is just 1 kg.  Spring constant k is unknown, but you can find it from given data:  

k = 0.75 N / 0.25 cm

= 3 N/cm, or 300 N/m.  

From the energy equation above, solve for v,

v = v √(k/m)  

= 0.15 √(300/1)

= 2.598 m/s.

b)  Momentum before impact = momentum after impact.

Since m = 1 kg,

v = 2.598 m/s,

p = 2.598 kg m/s.  

This is the same momentum carried by bullet as it strikes the block.  Therefore, if u is bullet speed,  

u = 2.598 kg m/s / 8 × 10⁻³ kg

= 324.76 m/s.

Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.

6 0
3 years ago
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