An object has an acceleration of 12.0 m/s/s. If the net force

It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft

Paul [167]

Answer: $1.23\ m/s^2$

Explanation:

Given

At an elevation of $y=34.7\ km$ , spacecraft is dropping vertically at a speed of $u=293\ m/s$

Final velocity of the spacecraft is $v=0$

using equation of motion i.e. $v^2-u^2=2as$

Insert the values

$\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2$

Therefore, magnitude of acceleration is $1.23\ m/s^2$ .

8 0

2 years ago

A vector is parallel to the y axis, what is it x component?<br>

slavikrds [6]

Answer:

Explanation:

A vector is parallel to the y axis .

Let its magnitude be A . So the vector can be represented as A j .

where i and j are unit vectors in x and y axis direction .

The x component of A j will be dot product of A j with i

The x component of A j = A j . i

= A x 0 [ Since j . i = 0 ]

= 0

4 0

2 years ago

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

NeX [460]

Answer:

The transverse displacement is $y(1.51 , 0.150) = 0.055 m$

Explanation:

From the question we are told that

The generally equation for the mechanical wave is

$y(x,t) = Acos (kx -wt)$

The speed of the transverse wave is $v = 8.25 \ m/s$

The amplitude of the transverse wave is $A = 5.50 *10^{-2} m$

The wavelength of the transverse wave is $\lambda = 0540 m$

At t= 0.150s , x = 1.51 m

The angular frequency of the wave is mathematically represented as

$w = vk$

Substituting values

$w = 8.25 * 11.64$

$w = 96.03 \ rad/s$

The propagation constant k is mathematically represented as

$k = \frac{2 \pi}{\lambda}$

Substituting values

$k = \frac{2 * 3.142}{0. 540}$

$k =11.64 m^{-1}$

Substituting values into the equation for mechanical waves

$y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))$

$y(1.51 , 0.150) = 0.055 m$

4 0

3 years ago

An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration

lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward = g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward = g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0

3 years ago

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0

3 years ago

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