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Digiron [165]
3 years ago
13

What of the following is an accurate statement

Physics
1 answer:
Valentin [98]3 years ago
5 0
B is right.  Because of the larger turns on the secondary the voltage on the secondary is larger.  This is what is meant by "step-up transformer"  The voltage is stepped up to a higher value
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A sound wave propagates through a region filled with an ideal gas at constant temperature T. It approaches an acoustically perme
nataly862011 [7]

Answer:

The answer is <em>e.2</em>

Explanation:

We should make use of Snell's refractive law. The arriving wave has a certain velocity at T in a medium, then instantly it reaches a medium (same composition) at T' where velocity would either decrease or increase.

When the incidence angle is 30 °, and we want to make the refraction angle 90 ° such that no sound passes through the barrier (this would be named total internal refraction), so we want the second medium to be "faster" than in the first.

<em>The steps are in the image attached:</em>

5 0
3 years ago
Determine the volume displaced and then calculate the density of this 54 g sample of brass.
inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density d_{w} of water is : 1000kgm^{-3}=1gcm^{-3}

Buoyant force exerted by a liquid on an object with V_{imm} of it's volume immersed is :

F_{B}=V_{imm}*d_{l}*g

where ,

  • F_{B} is the buoyant force
  • d_{l} is the density of the liquid
  • g is the acceleration due to gravity

Thus at equilibrium:

m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}

from these , we get the density of brass to be 1gcm^{-3}

which is not possible

7 0
3 years ago
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th
Alex Ar [27]

Answer:

 avriage force F = 2722.5 N

Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

      vf² = v₀² - 2 ax

The final carriage speed is zero (vf = 0)

      0 = v₀² - 2ax

      a = v₀² / 2x

      a = 1.1²/(2 0.200)

      a = 3.025 m / s²

      a = 3.0 m/s²

We calculate the average force

      F = ma

      F = 900 3,025

      F = 2722.5 N

3 0
3 years ago
2. Three blocks, A,B and C of mass 2kg. 3kg. 5kg respectively kept side by side with one another are accelerated at 2m/s2 across
gulaghasi [49]

Answer:

Total mass of combination = 2+3+5 = 10kg.

Acceleration produced = 2m/s^2

hence force =( total mass × acceleration)= (2×10)= 20 N.

Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N

applied force on 2 kg block = 20N

Force between 2 kg and 3 kg block = (20-4) = 16N. ans

Net force on 3 kg block = 3 × 2 =6N.

Applied force on 3 kg block due to 2 kg block = 16N.

hence, force between 3 kg and 5 kg block = (16-6) = 10N .

answers:-

(a) 20 N

(b) 16N

(c) 10 N

4 0
2 years ago
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