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3 years ago

2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table at a

Physics

1 answer:

elena55 [62]3 years ago

5 0

Answer:

Below

Explanation:

Here is what we are given in the problem :

dy = - 1.0 m (this is negative because the cat falls)

Vx = 5.0 m/s

We can use this equation to find the time that it will take the cat to fall :

t = √-2dy / g

t = √-2(-1.0) / 9.8

t = √0.20408

t = 0.45 (round this to 2 sig figs)

t = 0.5 seconds

Now we can find dx (distance in the x direction)

dx = vx (t)

dx = (5.0m/s)(0.45s)

dx = 2.258769 m

dx = 2.3 m

Hope this helps! Best of luck <3

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Anna71 [15]

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4 0

3 years ago

The graph represents the reaction 2H2 + 02 32H20 as it reaches

Alex

Answer:

C and D

Explanation:

5 0

3 years ago

Read 2 more answers

The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

7 0

4 years ago

Agent Bond is standing on a bridge, 17 m above the road below, and his pursuers are getting too close for comfort. He spots a fl

faust18 [17]

Answer:

2 poles

Explanation:

We are told Agent Bond is standing on a bridge, 17 m above the road below.

Now, the bed of the truck is 1.5 m above the road.

It means the distance hewill need to fall = 17 - 1.5 = 15.5 m

acceleration due to gravity = 9.8 m/s²

Initial velocity = 0 m/s

Thus, to find the time, we will use the equation;

s = ut + ½at²

Plugging in the relevant values gives;

15.5 = 0 + ½9.8t²

Multiply through by 2 to give;

15.5 × 2 = 9.8t²

31 = 9.8t²

t² = 31/9.8

t = √(31/9.8)

t = 1.78 sec

We are told that he spots a flatbed truck approaching at 28 m/s and that the telephone poles are 22m apart.

Thus; The truck will pass the poles at a rate of; Velocity/distance between poles = 28/22 = 1.273 poles per second

Since, we got the time to be 1.78 seconds, then we can find the number of poles by multiplying it with the rate of poles per seconds.

Thus;

Number of poles = 1.78 second × 1.273 poles/ seconds = 2.27 poles

This is approximately 2 poles

6 0

3 years ago

A truck goes 5 m/s to 25 in 20 seconds. What is the acceleration of the truck during 20-seconds interval?

Aliun [14]

Acceleration = (change in speed) / (time for the change)

Change in speed = (ending speed) - (starting speed)

Change in speed = (25 m/s) - (5 m/s) = 20 m/s

Time for the change = 20 sec

Acceleration = (20 m/s) / (20 sec)

<em>Acceleration = 1 m/s²</em>

4 0

3 years ago