A basic observation of a star is how bright it appears. This brightness is known as the star's

Stacy set up three vials on a hot plate. He poured the same amount of liquid

A because he is putting three vitals on a hot plate

4 0

3 years ago

9. A football punter attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in th

vichka [17]

Answer:

Angle is 55.52°

and Initial Speed is v=26.48 m/s

Explanation:

Given data

$x_{o}=0m\\ y_{o}=1.23m\\a_{oy}=a_{1y}=g=-9.8m/s^{2} \\x_{1}=67.0m\\y_{1}=0m\\t_{o}=0\\a_{ox}=m/s^{2} \\t_{1}=4.50s$

Applying the kinematics equations for motion with uniform acceleration in x and y direction

So

$x_{1}=x_{o}+v_{ox}t_{1}=67.0m\\0+4.50v_{o}Cos\alpha =67.0m\\v_{o}Cos\alpha =14.99\\v_{o}=14.99/Cos\alpha.....(1) \\and\\y_{1}=y_{o}+v_{oy}t_{1}+(1/2)a_{oy}t_{1}^{2} =0m\\ 1+4.50v_{o}Sin\alpha+(-9.8/2)(4.5)^{2}=0\\ v_{o}Sin\alpha=21.828.....(2)$

Put the value of v₀ from equation (1) to equation (2)

So

$\frac{14.99}{Cos\alpha }(Sin\alpha ) =21.828\\as\\tan\alpha =Sin\alpha /Cos\alpha \\So\\14.99tan\alpha =21.828\\tan\alpha =21.828/14.99\\\alpha =tan^{-1}(21.828/14.99) \\\alpha =55.52^{o}$

Put that angle in equation (1) or equation (2) to find the initial velocity

So from equation (1)

$v_{o}=(\frac{14.99}{Cos\alpha } ) \\v_{o}=(\frac{14.99}{Cos(55.52) } ) \\v_{o}=26.48m/s$

7 0

3 years ago

QUESTION 1 Linear Motion

Misha Larkins [42]

Answer:

48.51ms / 174.6 km/h

Explanation:

y = 1/2 x g x t^2 v = g x t

when y = 120m

120 = 1/2 x 9.8 x t^2

t^2 = 24.49

t = 4.95s

when t = 4.95s

v = 9.8 x 4.95

v = 48.51 m/s = 174.6 km/h

I'd say its realistic. But I don't really know that sry

3 0

2 years ago

An astronaut sitting on the launch pad on Earth's surface is 6,400 kilometers from Earth's center and weighs 400 newtons. Calcul

PIT_PIT [208]

Answer:

weight at height = 100 N .

Explanation:

The problem relates to variation of weight due to change in height .

Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

At the surface :

Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

400 = G Mm / (6400 x 10³ )²

G Mm = 400 x (6400 x 10³ )²

At height h from centre

mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²

= 400 / 4

= 100 N .

weight at height = 100 N

5 0

3 years ago

If a 6V battery is connected to a light bulb whose resistance is 55,000Ω How much current will flow in the circuit?

notka56 [123]

Answer:

Current, I = 0.000109 Amps

Explanation:

Given the following data;

Voltage = 6V

Resistance = 55,000 Ohms

To find the current flowing through the circuit;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

$V = IR$

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Making current the subject of formula, we have;

$I = \frac {V}{R}$

Substituting into the formula, we have;

$I = \frac {6}{55000}$

Current, I = 0.000109 Amps

5 0

3 years ago