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3 years ago

A 5.00 g projectile has a velocity of 255 m/s right. Find the force to stop this projectile in 1.45s.

Physics

1 answer:

Stells [14]3 years ago

8 0

Explanation:

The X-component of the velocity = Vcosx. Where, V = magnitude of the velocity. The x component of velocity will depend on the diagram. It the angle is measured from the x-axis which is considered the horizontal then Vx = Vcos(theta). The magnitudes of the components of velocity v → are v x = v cos θ and v y = v sin θ , v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure 4.12. Derivation of the Trajectory Formula.

y = refers to the vertical position of the object in meters. x = refers to the horizontal position of the object in meters. Horizontal velocity component: Vx = V * cos(α)

Vertical velocity component: Vy = V * sin(α)

Time of flight: t = [Vy + √(Vy² + 2 * g * h)] / g.

Range of the projectile: R = Vx * [Vy + √(Vy² + 2 * g * h)] / g.

Maximum height: hmax = h + Vy² / (2 * g)

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Murljashka [212]

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P = 10000 N/m²

F = P • A

F = (10000 N/m²)(1 m²)

F = 10000 N

At the side of the tank :

Pav = ½ρgH

Pav = ½(1000 kg/m³)(10 m/s²)(1 m)

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2 years ago

Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir

yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) $T=2\pi \sqrt{\frac{r^{3}}{GM}}$

$M=\frac{4\pi ^{2}r^{3}}{GT^{2}}$

$M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}$

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

$v=\frac{2\pi r}{T}$

$v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}$

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

$E = \frac{GM}{r^{2}}$

$E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}$

E = 0.28 N/kg

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3 years ago

Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direc

Lerok [7]

Answer:

424.26 m/s

Explanation:

Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q

The relative speed will be calculated by using pythagorean theorem

Relative speed = sqrt(300^2 + 300^2)

Relative speed = sqrt( 180000 )

Relative speed = 424.26 m/s

Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s

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3 years ago

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Natasha_Volkova [10]

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