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Arlecino [84]
3 years ago
13

When the flower below is rotated clockwise 90°, where will the red petal be in the image?

Physics
1 answer:
murzikaleks [220]3 years ago
6 0
Hey I need a pic then I'll answer in the comments. If no pic then if a red petal is directly facing a side, then it will be facing the adjacent clockwise side.
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PLEASE HELP. THIS IS DUE TODAY AT 11:59PM ​
OLga [1]

Answer:

ggn

Explanation:

rt

4 0
3 years ago
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on
g100num [7]

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

6 0
3 years ago
What are the horizontal and vertical components of a lizard’s displacement if it has climbed 7m directly up a tree?
madam [21]

Answer:

The horizontal component is zero.

The vertical component is 7\sin\theta

Explanation:

Given that,

The lizard climb 7m directly up on a tree.

We know that,

The horizontal component is

x=\cos\theta

The vertical component is

y=\sin\theta

If the lizard climb 7m directly up on a tree then,

We need to find the components

Using given data

The horizontal component of lizard is

x=0

The vertical component is

y=7\sin\theta

Hence, The horizontal component is zero.

The vertical component is 7\sin\theta

7 0
3 years ago
A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel
Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0
3 years ago
A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]
jarptica [38.1K]

-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.

             Speed  =  (original speed) + (gravity  x  falling time)

-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is

                         =       (0)            + (9.8 m/s²  x  3 sec)

                       Velocity  =  29.4 m/s downward . 
6 0
3 years ago
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