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SVETLANKA909090 [29]

2 years ago

13

You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g

. Take the origin at the launch point. Suppose that y-axis is directed upward and speed v0 is in the x-direction. How long after you launch the potato has it moved as far horizontally from the launch point as it has moved vertically

1 answer:

Ray Of Light [21]2 years ago

8 0

Answer:

$\displaystyle t=\frac{2v_o}{g}$

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

$\displaystyle y=\frac{gt^2}{2}$

The horizontal distance is

$x=v_ot$

We need to find the time when both distances are equal, thus

$\displaystyle \frac{gt^2}{2}=v_ot$

Simplifying by t

$\displaystyle \frac{gt}{2}=v_o$

Solving for t

$\displaystyle \boxed{t=\frac{2v_o}{g}}$

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A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is

astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B =(1.40 T)i

B =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

= - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C* 5.39* 10⁴ m/s* N/ C*m/s (-k)

F= 6.68*10¹¹⁴ N (-k)

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

=( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C* ( 5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴ i + 7.27*10¹¹⁴ j ) N

8 0

2 years ago

Many birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertica

scoray [572]

Answer:

0.109

Explanation:

8 0

2 years ago

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

Svetllana [295]

Answer:

The amount of charge the space shuttle collects is -1.224nC

Explanation:

The magnitude of Electric potential is given as;

V = kq/r

where;

V is the electric potential in volts

k is coulomb's constant

r is the radius of the sphere or distance moved by the charge

given; V = -1.1 V, k = 8.99 x 10⁹ Nm²/C², r = 10m

Substituting this values in the above equation, we estimate the amount of charge space shuttle collects.

q = (V*r)/k

q = (-1.1 *10)/(8.99 x 10⁹ )

q = -1.224 X 10⁻⁹ C

q = -1.224nC

Therefore, the amount of charge the space shuttle collects is -1.224nC

8 0

2 years ago

Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.

Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0

2 years ago

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

2 years ago