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hammer [34]
2 years ago
7

Sound travels through air at 343 m/s at 20 °C. A bat emits an ultrasonic squeak and hears the echo 0.05 second later . How far a

way was the object that reflected it?
Physics
1 answer:
lisabon 2012 [21]2 years ago
4 0

Answer:

8.575 m

Explanation:

speed = distance/time

distance -- 2d

343 = 2d/0.05

2d = 17.15

d = 17.15/2

d = 8.575

Thenks and mark me brainliest :))

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Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation
stira [4]

Answer:

Remains the same

Explanation:

5 0
2 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
(b) 11 - 3.72a 
(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
Now that you have the first derivative, it will give you the velocity as a function of t. 
(a) Velocity after 2 seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56 
So the velocity is 3.56 m/s  
(b) Velocity after a seconds. 
y' = 11 - 3.72t 
y' = 11 - 3.72a  
So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
y' = 11 - 3.72a
 y' = 11 - 3.72*5.913978495
 y' = 11 - 22
 y' = -11 
 So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
3 0
2 years ago
What is the moment of inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a
Paha777 [63]
E = mgh +  \frac{1}2} m v^{2} + \frac{1}{2} I \omega^{2} = mgh +  \frac{1}2} m  r ^{2}   \omega ^{2}  + \frac{1}{2} I \omega^{2}

for a solid cylinder:  I =  \frac{1}{2} m r^{2}
for a hollow cylinder: I = mr^{2}

I will look at the case of a hollow cylinder:

E = mgh + I \omega ^{2} = constant \\ \\ I =  \frac{mgh}{  \omega^{2} }

That is as far as i get.


7 0
3 years ago
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