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3 years ago

Just need help with 1 and 2 please :D i’m having a bit of trouble :/

Physics

1 answer:

dexar [7]3 years ago

7 0

1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.

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a length 650 cm of thin thread wraps around a cylinder exactly 30 times calculate the circumference and the radius of the cylind

Eduardwww [97]

Answer:

circumference= 65/3 cm = 21.67 cm

radius R = 3.45 cm

Explanation:

To calculate the length of the circumference of the cylinder, we divide 650 cm by 30 (the number of times it wrapped exactly around it)

length of circumference= 65/3 cm = 21.67 cm

now use the formula of the circumference length to find the radius (R):

circumference length = 2 * pi * R

65/3 = 2 * pi * R

R = 65 / (6 pi)

R = 3.45 cm

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3 years ago

A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator

cricket20 [7]

Answer:

a)= 98kJ

b)=108kJ

c) = 10kJ

Explanation:

a. The work that is done by gravity on the elevator is:

Work = force * distance

= mass * gravity * distance

= 1000 * 9.81 * 10

= 98,000 J

= 98kJ

b)The net force equation in the cable

T - mg = ma

T = m(g+a)

T = 1000(9.8 + 10)

T = 10800N

The work done by the cable is

W = T × d

= 10800N × 10

= 108000

=108kJ

c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J

Work done by cable = PE +KE

108,100 J = KE + 98,100 J

KE = 10,000 J

= 10kJ

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3 years ago

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A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

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3 years ago

Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc

zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

$Zn + CuSO_4 --> ZnSO_4 + Cu$

so here we have

$Zn = 10.1 g$

$CuSO_4 = 18.6 g$

$ZnSO_4 = 20 g$

$Cu = 8.7 g$

Now total mass of reactant is given as

$M_1 = 10.1 + 18.6 = 28.7 g$

Mass of the product is given as

$M_2 = 20 + 8.7 = 28.7 g$

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

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3 years ago

Give two conditions in which an object becomes weightless

Evgesh-ka [11]

Answer:

A body becomes weightless in a zero-gravity scenario and when a force is applied to a body that is equal and opposite to the force of gravity.

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3 years ago

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