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3 years ago

8

How does the vertical component of a pro- jectile’s motion compare with the motion of vertical free fall when air resistance is

negligi- ble?

1 answer:

Elena-2011 [213]3 years ago

7 0

Answer:

Explanation:

In case of projectile motion, the acceleration acts is in vertical direction only which is due to the acceleration due to gravity.

In case of free fall the body moves along the vertical straight line, which is again under the acceleration due to gravity. Thus the motion of a body under free fall and the vertical motion of a projectile is same.

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Is the distance traveled during a specific unit of

Anton [14]

Answer:

velocity

Explanation:

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4 years ago

What’s is the definition of STD

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S- sexually T- transmitted D- disease

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4 years ago

A spring (k=15.19kN/m)is is compresses 25cm and held in place on a 36.87° incline. A block (M=10kg) is placed on the spring. Whe

Savatey [412]

Answer:

The maximum vertical displacement is 2.07 meters.

Explanation:

We can solve this problem using energy. Since there is a frictional force acting on the block, we need to consider the work done by this force. So, the initial potential energy stored in the spring is transferred to the block and it starts to move upwards. Let's name the point at which the block leaves the ramp "1" and the highest point of its trajectory in the air "2". Then, we can say that:

$E_0=E_1\\\\U_e_0=K_1+U_g_1+W_f_1$

Where $U_e_0$ is the elastic potential energy stored in the spring, $K_1$ is the kinetic energy of the block at point 1, $U_g_1$ is the gravitational potential energy of the block at point 1, and $W_f_1$ is the work done by friction at point 1.

Now, rearranging the equation we obtain:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgh_1+\mu Ns_1$

Where $k$ is the spring constant, $x$ is the compression of the spring, $m$ is the mass of the block, $v_1$ is the speed at point 1, $g$ is the acceleration due to gravity, $h_1$ is the vertical height of the block at point 1, $\mu$ is the coefficient of kinetic friction, $N$ is the magnitude of the normal force and $s_1$ is the displacement of the block along the ramp to point 1.

Since the force is in an inclined plane, the normal force is equal to:

$N=mg\cos\theta$

Where $\theta$ is the angle of the ramp.

We can find the height $h_1$ using trigonometry:

$h_1=s_1\sin\theta$

Then, our equation becomes:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv_1^{2}+mgs_1\sin\theta+\mu mgs_1\cos\theta\\\\\implies v_1=\sqrt{\frac{2(\frac{1}{2}kx^{2}-mgs_1\sin\theta-\mu mgs_1\cos\theta)}{m}}=\sqrt{\frac{kx^{2}}{m}-2gs_1(\sin\theta+\mu \cos\theta)}$

Plugging in the known values, we get:

$v_1=\sqrt{\frac{(15190N/m)(0.25m)^{2}}{10kg}-2(9.8m/s^{2})(1.12m)(\sin36.87\°+(0.300) \cos36.87\°)}\\\\v_1=8.75m/s$

Now, we can obtain the height from point 1 to point 2 using the kinematics equations. We care about the vertical axis, so first we calculate the vertical component of the velocity at point 1:

$v_1_y=v_1\sin\theta=(8.75m/s)\sin36.87\°=5.25m/s$

Now, we have:

$y=\frac{v_1_y^{2}}{2g}\\\\y=\frac{(5.25m/s)^{2}}{2(9.8m/s^{2})}\\\\y=1.40m$

Finally, the maximum vertical displacement $h_2$ is equal to the height $h_1$ plus the vertical displacement $y$ :

$h_2=h_1+y=s_1\sin\theta +y\\\\h_2=(1.12m)\sin36.87\°+1.40m\\\\h_2=2.07m$

It means that the maximum vertical displacement of the block after it becomes airborne is 2.07 meters.

7 0

3 years ago

Having aced your introductory physics course, you are hired as a summer intern at NASA. You are sent as part of a team to explor

djverab [1.8K]

Answer:

1) M = 9.7*10^21 kg

2, 3) 1610.81 m/s

4) 312.911 km

5) 1315.65 m/s

Explanation:

Given:-

- The radius of the asteroid, R = 499 km

- The surface acceleration, g = 2.6 m/s^2

Solution:-

- To determine the mass of the asteroid ( M ) we will use the relation for the gravitational acceleration produced by a spherical object of mass ( M ) as follows:

$g = G\frac{M}{R^2}$

Where,

G: the universal gravitational constant = 6.674*10^−11

- Use the radius of asteroid ( R ) and the given surface acceleration ( g ) and solve for the mass ( M ) of the asteroid using the relation given above:

$M = \frac{gR^2}{G} \\\\M = \frac{2.6*(499,000)^2}{6.674*10^-^1^1} \\\\M = 9.7*10^2^1 kg$

- To determine the escape velocity of the mass of rock ( m = 4kg ) from the gravitational pull of the asteroid. We will use the conservation of energy principle.

- The conservation of energy principle states:

$K.E_i + P.E_i = K.E_f + P.E_f$

Where,

K.Ei: The initial kinetic energy of the rock of mass ( m )

P.Ei: The potential energy of the system at the surface

K.Ef: The final kinetic energy of the rock of mass ( m )

P.Ef: The final potential energy of the system at infinite

- The gravitational potential energy of the system of an object of mass (m ) at any distance ( r ) from the center of the more massive body ( M ) is given as:

$P.E = - G\frac{M*m}{r}$

- The escape velocity is just enough initial velocity ( ve ) that allows an object to cross the gravitational effect of the massive body. Once the effect of the gravity is insignificantly small ( infinite ). Almost all of the kinetic energy has been lost by doing work against the gravitational pull.

- Therefore, K.Ef = P.Ef = 0 ( at infinity ).

$0.5mv_e^2 - G\frac{Mm}{r} = 0\\\\v_e = \sqrt{2G\frac{M}{r} } = \sqrt{2G\frac{M}{R} } \\\\v_e = \sqrt{2(6.674*10^-^1^1)\frac{9.7*10^2^1}{499,000} }\\\\v_e = 1610.81 \frac{m}{s}$

- From the above relationship derived for an object at the surface of an asteroid body to escape the grasp of the gravitational pull is independent of the mass of the object ( m ). Hence, whatever the mass of the object is it does not affect the required escape velocity.

Answer: The rock of mass m = 8 kg and m = 4 kg require minimum vertical velocity of 1610.81 m/s to escape the asteroid gravitational pull.

- For the final part we will again apply the principle of conservation of energy for the system of asteroid of mass ( M ) and an object of mass ( m ). We have:

$K.E_i + P.E_i = K.E_f + P.E_f$

Where,

K.Ei: The initial kinetic energy of the rock of mass ( m ) = 0 ( dropped)

P.Ei: The potential energy of the system at altitude ( h )

K.Ef: The final kinetic energy of the rock of mass ( m ) at surface

P.Ef: The final potential energy of the system at surface

$0 - G\frac{Mm}{( R + h )} = 0.5mv^2 - G\frac{Mm}{ R } \\\\v^2 = 2GM* [ \frac{1}{R} - \frac{1}{R+h} ] \\\\v = \sqrt{2(6.674*10^-^1^1)*(9.70*10^2^1)* [ \frac{1}{499,000} - \frac{1}{1499000}]}\\\\v = 1315.65 \frac{m}{s}$

Answer: The speed of the 8kg rock upon hitting the surface of asteroid would be 1315.65 m/s

- Similarly, how far away a rock of mass ( m ) can go to from the surface of asteroid if it leaves the surface with initial velocity vi = 1000 m/s. We will use the energy conservation expression derived in the previous part. We have:

$0.5mv^2 - G\frac{Mm}{R} = G\frac{Mm}{R + h} \\\\\frac{0.5v^2}{GM} - \frac{1}{R} = -\frac{1}{R + h}\\\\R + h = \frac{1}{-\frac{0.5v^2}{GM} + \frac{1}{R}} \\\\h = \frac{1}{-\frac{0.5v^2}{GM} + \frac{1}{R}} - R\\\\h = \frac{1}{-\frac{0.5(1000)^2}{(6.674*10^-^1^1)*(9.7*10^2^1)} + \frac{1}{(499000)}} - 499000\\\\h = 312.911 km$

Answer: The 4 kg rock would be able to travel 312.911 km above the asteroid surface if it had an initial velocity of 1000 m/s.

8 0

3 years ago

A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant vel

omeli [17]

Answer:

Explanation:

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate .

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration . It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N .

3 0

4 years ago