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3 years ago

Would someone just answer this now?

Physics

1 answer:

sineoko [7]3 years ago

3 0

Angle = 90 - 39.4 = 50.6
cos50.6 = adjacent/47.3
(47.3)(cos50.6) = 30.02

The y component of the vector is 30.0 m.

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Where does a phase change occur

il63 [147K]

A phase change is occuring; the liquid water is changing to gaseous water, or steam. On a molecular level, the intermolecular forces between the water molecules are decreasing. The heat is providing enough energy for the water molecules to overcome these attractive forces.

3 0

3 years ago

Read 2 more answers

An object located near the surface of Earth has a weight of a 245 N

marusya05 [52]

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

$m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg$

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

4 0

3 years ago

Aristotle believed which of the following to be true?

Bumek [7]

Answer:

number 2

Explanation:

5 0

3 years ago

Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel

coldgirl [10]

Answer: from the vertical, one should aim 86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² + 1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ into equ 2

365.76(cos∅) × 5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ = 3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0

4 years ago

Determine the volume displaced and then calculate the density of this 54 g sample of brass.

inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density $d_{w}$ of water is : $1000kgm^{-3}=1gcm^{-3}$

Buoyant force exerted by a liquid on an object with $V_{imm}$ of it's volume immersed is :

$F_{B}=V_{imm}*d_{l}*g$

where ,

$F_{B}$ is the buoyant force
$d_{l}$ is the density of the liquid
$g$ is the acceleration due to gravity

Thus at equilibrium:

$m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}$

from these , we get the density of brass to be $1gcm^{-3}$

which is not possible

7 0

3 years ago