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1 year ago

Which of the following would be the magnitude of the vector given a horizontal component of 30 and a

Physics

1 answer:

Scorpion4ik [409]1 year ago

7 0

Answer:

Explanation:

Use the Pythagorean theorem to find the length of the diagonal, or the hypotenuse of an imaginary triangle. 30^2 + 40^2 = 2500, which is 50^2. So, the magnitude is 50.

Brainliest, please :)

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A toy rocket is fired vertically into the air from the ground at an initial velocity of 80 feet per second. Find the time it wil

miskamm [114]

Answer: 16.3 seconds

Explanation: Given that the

Initial velocity U = 80 ft/s

Let's first calculate the maximum height reached by using third equation of motion.

V^2 = U^2 - 2gH

Where V = final velocity and H = maximum height.

Since the toy is moving against the gravity, g will be negative.

At maximum height, V = 0

0 = 80^2 - 2 × 9.81 × H

6400 = 19.62H

H = 6400/19.62

H = 326.2

Let's us second equation of motion to find time.

H = Ut - 1/2gt^2

Let assume that the ball is dropped from the maximum height. Then,

U = 0. The equation will be reduced to

H = 1/2gt^2

326.2 = 1/2 × 9.81 × t^2

326.2 = 4.905t^2

t^2 = 326.2/4.905

t = sqrt( 66.5 )

t = 8.15 seconds

The time it will take for the rocket to return to ground level will be 2t.

That is, 2 × 8.15 = 16.3 seconds

8 0

3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago

True or false the hotter the star the higher is absolute magnitude?

liubo4ka [24]

Answer:

true

Explanation:

6 0

3 years ago

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

2 years ago

An object is most likely to sink in water if

LiRa [457]

Answer:

High density D answers to your questions

8 0

2 years ago

Read 2 more answers