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2 years ago

Differentiate the following functions with respect to x xsin x

Physics

2 answers:

gtnhenbr [62]2 years ago

6 0

Answer:

Here is your answer

Hope it helps

Vikki [24]2 years ago

3 0

The answer is xcos x + sin x

Explanation:

d/dx (xsin x)

= x d/dx (sin x) + sin x d/dx (x)

= xcos x + sin x (1)

= xcos x + sin x

Thus, The answer is xcos x + sin x

<u>-TheUnknownScientist</u>

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Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

gregori [183]

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

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2 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

2 years ago

What effect does the vertical acceleration have on the horizontal velocity of the projectile?

KengaRu [80]

Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity..
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.

3 0

2 years ago

An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m

Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0

2 years ago

How did the astronomy of hipparchus and ptolemy violate the principles of early greek philosophers?

Mariana [72]

Answer: Hipparchus & Ptolemy believed sun and moon travel around circles with Earth in the middle (geocentric) while Plato and Aristotle believed Perfect unchanging heavens

Explanation: thus hipparchus and ptolemy violate early greek philosophy

8 0

2 years ago

Differentiate the following functions with respect to x xsin x​

Differentiate the following functions with respect to x xsin x