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3 years ago

7

The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440 × 103 m), and its mass is 3 × 1023 kg. An

object is launched straight up from just above the atmosphere of Mercury.
a. What initial speed is needed so that when the object is far from Mercury its final speed is 2000 m/s?
b. What initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s?

1 answer:

KonstantinChe [14]3 years ago

5 0

Answer:

(a)4676m/s

(b) 4227m/s

Explanation:

Let the mass of mercury be Mm = 3×10²³ kg and the mass of the object be M

Also let the distance from the center of mercury be R1 = 2.240×10⁶m let the distance the object is launched to be R

From the energy conservation equation

K1 + U1 = K2 + U2

K1, K2 = initial and final kinetic energies.

U1, U2 = initial and final potential energies.

1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)

Assuming that the very far R approaches infinity then the second term to the right side of the equation above approaches zero

So dividing through by M since M is common to both sides we have that

1/2V1² + (-G×Mm/R1) = 1/2V2²

Rearranging,

V1² = 2×( 1/2V2²+ GMm/R1 )

Given V2 = 2000m/s and also the constant value of G = 6.67×10‐¹¹ N•m²/kg²

Substituting,

V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))

V1² = 21866071.43

V1 = 4676m/s

(b) this is still similar to the case described above but this time V2 = 0m/s

V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1

V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)

= 17866071.43

V1 = 4227m/s

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Marta_Voda [28]

a) The momentum of the coconut is 3 kg m/s

b) At first, the air resistance is negligible, so the coconut accelerates due to the force of gravity

c) The coconut reaches its terminal velocity

Explanation:

a)

The momentum of an object is given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

For the coconut in this problem, we have:

m = 1.5 kg (mass)

v = 2 m/s (velocity)

Therefore, its momentum is

$p=(1.5)(2)=3 kg m/s$

B)

There are only two forces acting on the coconut during its fall:

The force of gravity, of magnitude $mg$ (m= mass of the coconut, g = acceleration of gravity), acting downward
The air resistance, acting upward, whose magnitude is proportional to the speed of the coconut

During the first momentums of the fall, the speed of the coconut is still low, so the air resistance is mostly negligible, and therefore only the force of gravity is acting on the coconut. Since this force is constant, it means that the acceleration of the coconut is constant: therefore, its velocity keeps increasing during the fall, and the coconut speeds up.

C)

If the tree is very tall, the fall of the coconut lasts long, and the speed of the coconut keeps increasing. Since the air resistance is proportional to the speed, this means that at some point, the air resistance is no longer negligible, and it starts to have some effect on the fall of the coconut. In particular, at a certain point, the air resistance will become equal (in magnitude) to the force of gravity (but opposite in direction): this means that from this point, the acceleration of the coconut will be zero, and therefore the coconut will continue its motion at constant velocity. This velocity is called terminal velocity, and it occurs when the force of gravity is equal to the air resistance:

$mg = F_r$

where $F_r$ is the air resistance.

Learn more about forces and weight:

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8 0

2 years ago

NEED ANSWER ASAP WILL MARK BRAINLIEST

lukranit [14]

Speed=30 m/s - 1.5 m/s = 28.5 m/s forward

7 0

3 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

7 0

3 years ago

The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

<u>Given:</u>

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$

Thus, the y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

3 0

2 years ago