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djverab [1.8K]
2 years ago
12

A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t

ime between the person see and hear a firework explosion
Physics
1 answer:
WINSTONCH [101]2 years ago
3 0

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s

So, the required time is 2.2 seconds.

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A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top?
kati45 [8]

Answer:

8361.063 J

Explanation:

7 0
2 years ago
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm
Genrish500 [490]

Answer:

5.95 A

Explanation:

From the question

R = ρL/A..................... Equation 1

Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.

Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m

Substitute these values into equation 1

R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶

R = 0.084 Ω.

Finally, using Ohm law,

V = IR

Where V = Voltage, I = current

Make I the subject of the equation

I = V/R............... Equation 2

I = 0.5/0.084

I = 5.95 A

4 0
3 years ago
A positive charge moving up enters a magnetic field pointing out of the screen. What is the direction of the magnetic force on t
erma4kov [3.2K]
According to the right-hand thumb rule, the forefinger gives the velocity of charge, the thumb gives the magnetic force and the center finger gives the direction of magnetic field.

then, as shown in the picture, the <span>direction of the magnetic force on the charge is in the right direction.</span>

6 0
3 years ago
Read 2 more answers
A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f
meriva

Answer:

v=\sqrt{\dfrac{10g(R-r)}{7}}

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

I=\dfrac{2}{5}mr^2

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for  rolling without slipping       v= ωr

Form energy conservation

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

 v= ωr

I=\dfrac{2}{5}mr^2

mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2

mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2

2mg(R-r)=mv^2+\dfrac{2}{5}mv^2

2g(R-r)=\dfrac{7}{5}v^2

v=\sqrt{\dfrac{10g(R-r)}{7}}

3 0
3 years ago
What equation gives the position at a specific time for an object with constant acceleration? 
malfutka [58]
Given constant acceleration, we can get the final position of an object in terms of both its initial velocity and its acceleration using one of the equations of motion.

The equation that we will use is:
Xf = Xi +  Vi*t + (1/2)*a*t^2
where:
Xf is the final position of the object
Xi is the initial position of the object
Vi is the initial velocity of the object
t is the time
a is the constant given acceleration
4 0
3 years ago
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