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3 years ago

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"the student drops one cannonball, and exactly 1.0 s later drops the other cannonball from the same height. what is the time int

erval between the first cannonball striking the ground and the second cannonball striking the ground"

1 answer:

Elden [556K]3 years ago

6 0

Answer;

1 second

Explanation;

Two objects moving at the same speed will always stay the same distance apart. If two objects are moving at different speeds, the distance between them must change.

Therefore; if the distance will be the same and the speed is also the same then the time taken will be the same.

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At the race track, one race car starts its engine with a resulting intensity level of 104.0 dB at point P. Then 6 more cars star

dem82 [27]

To solve the problem it is necessary to apply the concepts related to sound intensity. The most common approach to sound intensity measurement is to use the decibel scale:

$\beta (dB) = 10log_{10}(\frac{I}{I_0})$

Where,

$I_0 = 1*10^{-12}$ is a reference intensity. It is the lowest or threshold intensity of sound a person with normal hearing can perceive at a frequency of 1000 Hz.

I = Sound intensity

Our values are given by,

$\beta = 104dB$

$\#Autos = 7$

For each auto the intensity would be,

$104 = 10log\frac{I}{1*10^{-12}}$

$10.4= log_{10} (\frac{I}{10^{-12}})$

$10^{10.4}*10^{-12}=I$

$I = 0.02511W/m^2$

Therefore the sound intesity for the 7 autos is

$I= 7* 0.02511$

$I = 0.1748W/m^2$

The sound level for the 7 cars in dB is

$\beta (dB) = 10log_{10}(\frac{0.1748}{1*10^{-12}})$

$\beta (dB) = 112.42dB$

8 0

3 years ago

A baseball pitcher throws the ball towards the batter at 90 mph. His bat connects with the ball for a line drive, after which th

forsale [732]

Answer:

$F=-18412.9N$ , where the minus indicates the direction is opposite to that of the throw.

Explanation:

a)

Since MKS stands for meter-kilogram-second and we know that:

$1\ hour = 3600\ seconds$

$1\ mile = 1600\ meters$

$1000g = 1kg$

We can write that:

$\frac{1\ hour}{3600\ seconds}=1$

$\frac{1600\ meters}{1\ mile}=1$

$\frac{1kg}{1000g}=1$

These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.

So we have that:

$90 mph=90 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=40m/s$

$110 mph=110 \frac{miles}{hour}(\frac{1\ hour}{3600\ seconds})(\frac{1600\ meters}{1\ mile})=48.89m/s$

$145 g=145 g(\frac{1kg}{1000g})=0.145kg$

b)

Newton's 2nd Law tells us that F=ma, and the definition of acceleration is $a=\frac{\Delta v}{\Delta t}$ , so we have:

$F=m\frac{\Delta v}{\Delta t}=m\frac{v_f-v_i}{t}$

Taking the throw direction as the positive one, for our values we have:

$F=m\frac{v_f-v_i}{t}=(0.145kg)\frac{(-48.89m/s)-(+40m/s)}{0.0007s}=-18412.9N$

4 0

3 years ago

Before the positive psychology movement, psychology focused mainly on

KonstantinChe [14]

Before positive psychology, the main focus was mostly pathology, that is, studying various psychological issues and sometimes finding ways to treat them, or sometimes not and just studying them for the sake of studying them and noting their occurrence.

7 0

3 years ago

Read 2 more answers

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp

IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0

3 years ago

Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf

Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

$E_{m} = E_{k} + E_{p}$

$E_{m} = \frac{1}{2}v^{2} + gh$

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

$E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg$

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

$P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW$

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0

3 years ago