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kkurt [141]
3 years ago
9

With a velocity of 15 m/s, how long would it take the dog to run 100 meters?

Physics
2 answers:
Tcecarenko [31]3 years ago
4 0

Answer:

6.67 s

Explanation:

Assuming that the velocity is constant, then the governing equation in this case is:

Distance = Velocity x Time

or

Time = Distance / Velocity

= 100 / 15

= 6.67 s

Zinaida [17]3 years ago
3 0

Answer

6.66 seconds imma seem like an idiot if something like this is werong

Explanation:

dogg is speed

100/15

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What is Physics? How is Physics used in your everyday life? Give at least two examples and explain your answer. You need at leas
ikadub [295]

Answers:

The transportation industry is no stranger to the manipulation of everyday physics. Cars and trains utilize the wheel, which provides a smooth, steady motion.

The ears hear sounds which occur through the movement of air molecules, and the chemistry that drives all of biology depends on the physics of energy and molecules. Every day, for example, plants absorb sunlight, water, and carbon dioxide, creating glucose and releasing oxygen as a byproduct.

Brainlist pls!

3 0
3 years ago
How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),
fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice =  37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat  absorbed during this phase change be Q₂ .

Let heat  absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0
3 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
Microwave transmission involves sending signals from one microwave station to another. What is this often called?
astraxan [27]
It is called fixed wireless

4 0
3 years ago
A body is thrown up with a velocity of 78.4 m per second.How high will it rise and how much time it will take to return to its p
a_sh-v [17]

Answer:

The maximum height reached by the body is 313.6 m

The time to return to its point of projection is 8 s.

Explanation:

Given;

initial velocity of the body, u = 78.4 m/s

at maximum height (h) the final velocity of the body (v) = 0

The following equation is applied to determine the maximum height reached by the body;

v² = u² - 2gh

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (78.4²) / (2 x 9.8)

h = 313.6 m

The time to return to its point of projection is calculated as follows;

at maximum height, the final velocity becomes the initial velocity = 0

h = v + ¹/₂gt²

h = 0 + ¹/₂gt²

h =  ¹/₂gt²

2h = gt²

t² = 2h/g

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 313.6}{9.8} }\\\\t = 8 \ s

4 0
3 years ago
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