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Mkey [24]
3 years ago
6

PLEASE HELP ASAP

Physics
1 answer:
77julia77 [94]3 years ago
8 0

Answer:

<em>The first choice (32m/s) is the closest to the answer</em>

Explanation:

The magnitude of a vector is the distance between the initial and the end point of the vector.

Being Vx and Vy the horizontal and vertical components of the vector V respectively, the magnitude of V is calculated as:

\mid \vec{V} \mid =\sqrt{V_x^2+V_y^2}

The components of the velocity of the physics student's projectile launcher are Vx=28 m/s and Vy=15 m/s.

Calculate the magnitude of the velocity:

\mid \vec{V} \mid =\sqrt{28^2+15^2}

\mid \vec{V} \mid =\sqrt{784+225}

\mid \vec{V} \mid =\sqrt{1009}

\mid \vec{V} \mid =31.8 \ m/s

The first choice (32m/s) is the closest to the answer

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In a powder-flavored drink solution, which substance is the solvent?
AleksandrR [38]

Answer:

c. Water

Explanation:

The solvent is the substance in which a solute is dissolved to make the solution.

In this case, the water is the solvent, the flavored powder is the solute, and the solution is the final drink.

7 0
3 years ago
You move to Mars and as a momento from Earth take your Ma-maw's mercury barometer. You place it outside in the Martian atmospher
oksano4ka [1.4K]

Answer:

Explanation:

No, the reading is not expected to be accurate. This is because Relative to Earth, the air on Mars is extremely thin. The Martian atmosphere is primarily carbon dioxide with a much lower surface pressure, and Mars does not have oceans and an Earthlike hydrological cycle so latent heat release is not as important as it is for Earth.

8 0
3 years ago
hydrogen does not have the typical properties of a metal, However, hydrogen is located above group 1A because it has one______.?
frez [133]
One protons and also it grouped their because how it reacts with other materials such as other substances or materials
4 0
3 years ago
If Q = 16 nC, a = 3.0 m, and b = 4.0 m, what is the magnitude of the electric field at point P?
Lynna [10]
In BPC

tan\theta =a/b = 3/4

\theta = tan^-1(0.75)

\theta = 36.87 deg

BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m

Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C

Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C

Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C

Net electric field along X-direction is given as

Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C

Net electric field along X-direction is given as

Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C

Net electric field is given as

E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
8 0
3 years ago
PLEASE HELP
Gennadij [26K]

Answer:

3.675 m

Explanation:

a_{x} =0 v_{xo}=100 a_{y} =-g  v_{yo}=0

X-direction     | Y-direction

R=x_{o}+ v_{xo} t  | y=y_{o}+v_{yo}t+\frac{1}{2}a_{y}t^2

75=100t         |y=0+0+\frac{1}{2} (9.8)(0.75)

\frac{75}{100} =t             | y=3.675 m

0.75s=t              

Hope it helps

3 0
3 years ago
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