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aleksley [76]
3 years ago
11

Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. The length of the

Physics
1 answer:
Illusion [34]3 years ago
4 0

Answer:

\left \| \vec R \right \|=50\ inches

Explanation:

<u>Vectors In The Plane</u>

Given two vectors \vec X and \vec Y, their sum is shown as the vector \vec R in the image below. It can be seen the magnitude of R is the hypotenuse of a right triangle. i.e.

\left \| \vec R \right \|=\sqrt{\left \| X \right \|^2+\left \| Y \right \|^2}

The magnitude of \vec X is 48 inches and the magnitude of \vec Y is 14 inches, the magnitude of R is

\left \| \vec R \right \|=\sqrt{48^2+14^2}

\left \| \vec R \right \|=\sqrt{2304+196}=\sqrt{2500}

\left \| \vec R \right \|=50\ inches

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A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal
sammy [17]

Answer:

<h3>1.03684m</h3>

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0
3 years ago
A helium-filled balloon occupies a volume of 15 cubic meters at sea level. the balloon is released and raises to a point in the
Elena L [17]

According to Boyle’s law, For a fixed amount of an ideal gas kept at a fixed temperature, P (pressure) and V (volume) are inversely proportional.

Therefore,

P_{1} V_{1} =P_{2} V_{2}

Given P_{1} = 1 atm , V_{1} = 15 \ cubic \ meter and P_{2} = 0.75\ atm.

Thus,

V_{2} = \frac{P_{1}\times V_{1}  }{ P_{2} } = \frac{1\times 15}{0.75} =20 m^3

5 0
3 years ago
Why do the passengers on a high-flying airplane not appear weightless, similar to the astronauts on the space station?
sergey [27]

<span>Even in space, there is still presence of gravity. The cause of weightlessness is not how far above the earth the space shuttle is but rather how fast it is travelling. The shuttle is in free fall causing weightlessness, but it is travelling fast enough to miss the earth as it falls. Similarly, the airplane could also provide weightlessness if it went free fall as well. However, that ends as the plane hits the ground. </span>

4 0
3 years ago
Can anyone help me with this question please​
JulsSmile [24]

Explanation:

V=u+at

where,

v=final speed

u=initial speed,(starting speed)

a=acceleration

t=time

  1. v=u+at = 6=2+a*2

6=2+2a

2a=6-2

2a=4

a=4/2 = 2

a =2

2. to find time taken

v=u+at

25=5*2t

2t=25-5

2t=20

t=20/2

t=10sec

3. finding final speed

v=u+at

v=4+10*2

=4+20

v=24m/sec

5.v=u+at

=5+8*10

=5+80

V=85m/sev

6. v=u+at

8=u+4*2

8=u+8

U=8/8

u=1

these are your missing values

5 0
3 years ago
In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, an
Tju [1.3M]
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
8 0
4 years ago
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