Pics pls you know what I mean

Which sound wave-object interaction is used by animals in echolocation?

tekilochka [14]

The answer is reflection

5 0

3 years ago

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A car with a mass of 1,500 kg is accelerating at a rate of

AURORKA [14]

Answer:

The net force acting on the car is

3

×

10

3

Newtons.

Hope this helps you

Explanation:

Force is defined as the product of the mass of the body and its aaceleration,

⇒

F

=

m

a

Substituting the above given values we get,

F

=

(

1500

k

g

)

(

2.0

m

/

s

2

)

=

3000

N

=

3

×

10

3

N

.

4 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

how much time would it take for the sound of a thunder to travel 1,500 meters if sound travelers at a speed of 330 m/sec?

inysia [295]

(1,500 meters) x (1 sec/330 meters) =

(1,500 / 330) (meters-sec/meters) =

4.55 seconds

8 0

3 years ago

You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.

Blizzard [7]

Answer:

Remains same

Explanation:

$T$ = Time period of oscillation

$m$ = mass

$k$ = spring constant

Time period of oscillation is given as

$T = 2\pi \sqrt{\frac{m}{k} }$

we know that as we move from earth to moon, the value of spring constant "k" and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0

3 years ago