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3 years ago

Two ice skaters are at rest and facing each other. In order to move, they push

Physics

2 answers:

hoa [83]3 years ago

4 0

Answer C - the final total momentum must equal zero.

Paha777 [63]3 years ago

3 0

Answer:

Explanation:

Hopefully this helps

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A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of

IrinaK [193]

Answer:

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$ no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$

7 0

4 years ago

A vessel is filled with a liquid of density 1900 kg/m3 . There are two holes (one above the other) in the side of the vessel. Li

FromTheMoon [43]

Answer:

110 meters is the distance where they will intersect

Explanation:

given,

liquid density = 1900 kg/m³

distance of upper hole = 19 m

distance of lower hole = 117 m

acceleration due to gravity = 9.8 m/s²

the speed at each point

$v = \sqrt{2gh}$

for upper hole $v = \sqrt{2\times 9.8 \times 19}$

v = 19.29 m/s

lower hole $v = \sqrt{2\times 9.8 \times 117}$

v = 47.88 m/s

The path for each is parabolic

x = v t

$y = \dfrac{1}{2}gt^2$

$y = \dfrac{1}{2}g(\dfrac{x}{v})^2$

$y = \dfrac{gx^2}{2v^2}$

we get

upper hole

$y = \dfrac{9.8\times x^2}{2\times 19.29^2}= 0.0132 x^2$ lower hole

$y= \dfrac{9.8\times x^2}{2\times 47.88^2}=0.00214x^2$ y for upper hole = 80 + y for lower hole

$0.0132 x^2= 98 + 0.00214 x^2$

$0.0082 x^2 = 98$

x = 109.32 meters

110 meters is the distance where they will intersect

4 0

3 years ago

A spherical mirror gives an image magnified 5 times at a distance 5 m. determine whether the mirror is convex or concave? How mu

irakobra [83]

Answer:

1. Concave mirror.

2. 4.17 m or 417 cm.

Explanation:

The following data were obtained from the question:

Object distance (u) = 5 m

Magnification (M) = 5

Focal length (f) =..?

1. Identification of the mirror.

To determine whether or not the mirror is concave or convex, we must first of all calculate the image distance.

This can be obtained as follow:

Object distance (u) = 5 m

Magnification (M) = 5

Image distance (v) =.?

Magnification (M) = image distance (v) /object distance (u).

M = v/u

5 = v/5

Cross multiply

v = 5 x 5

v = 25 m

Since the image distance obtained is positive, the mirror is said to be a concave mirror.

2. Determination of the focal length of the mirro.

This can be obtained as follow:

Object distance (u) = 5 m

Image distance (v) = 25 m

Focal length (f) =...?

1/f = 1/v + 1/u

1/f = 1/25 + 1/5

1/f = 0.04 + 0.2

1/f = 0.24

Cross multiply

f x 0.24 = 1

Divide both side by 0.24

f = 1/0.24

f = 4.17 m

Converting the focal length of cm, we have:

1 m = 100 cm

Therefore, 4.17 m = 4.17 x 100 = 417 cm

Therefore, the focal length of the mirror is 4.17 or 417 cm.

4 0

3 years ago

PLEASE HELP ASAP!!!!

sveticcg [70]

Answer:

it's frequency

Explanation:

hope this helps :)

8 0

3 years ago

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Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 10^7 meters from the center of Earth. (The value of

Kryger [21]

Answer:

A. 2.36

Explanation:

This is the right answer for Plato users!

3 0

3 years ago

Read 2 more answers