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2 years ago

7

Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to

ok her 15 minutes to get home, what was the bus' average velocity during the return trip?

1 answer:

Dmitriy789 [7]2 years ago

7 0

Answer:

Explanation: idk sry

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The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An

Xelga [282]

Answer:

Explanation:

mass of probe m = 474 Kg

initial speed u = 275 m /s

force acting on it F = 5.6 x 10⁻² N

displacement s = 2.42 x 10⁹ m

A )

initial kinetic energy = 1/2 m u² , m is mass of probe.

= .5 x 474 x 275²

= 17923125 J

B )

work done by engine

= force x displacement

= 5.6 x 10⁻² x 2.42 x 10⁹

= 13.55 x 10⁷ J

C ) Final kinetic energy

= Initial K E + work done by force on it

= 17923125 +13.55 x 10⁷

= 1.79 x 10⁷ + 13.55 x 10⁷

= 15.34 x 10⁷ J

D ) If v be its velocity

1/2 m v² = 15.34 x 10⁷

1/2 x 474 x v² = 15.34 x 10⁷

v² = 64.72 x 10⁴

v = 8.04 x 10² m /s

= 804 m /s

5 0

2 years ago

Find the value of F1 + F2 + F3.<br>

Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u /> $F_{y}=2[N]$ <u />

<u>For F₂</u>

$F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]$

<u>For F₃</u>

<u /> $F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]$ <u />

Now we can sum each one of the forces in the given axes:

$F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]$

Now using the Pythagorean theorem we can find the total force.

$F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]$

8 0

2 years ago

How to use kinetic energy to calculate velocity

BartSMP [9]

As we know the formula of kinetic energy is

$KE = \frac{1}{2} mv^2$

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

$150,000 = \frac{1}{2} * 120 * v^2$

$v^2 = 2500$

$v = 50 m/s$

So speed of above object is 50 m/s

7 0

2 years ago

Which answer explains why plate B is moving underneath plate A

blsea [12.9K]

Answer:

A.

Explanation:

Earth is composed of different layers and one layer moves over another due to differences in the densities.

According to the physics of density, a substance having less density floats over a higher density substance. The oceanic crust has more density than the continental crust that is why continental crust float over oceanic crust.

So in the given example, plate B is moving below the plate A, it means plate B is more dense than plate A because plate B is composed of oceanic crust . <u>For example : continents float over the asthenosphere (a layer below the lithosphere).</u>

Hence, the correct answer is "A ".

4 0

2 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

Read 2 more answers