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3 years ago

A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL

Physics

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

0.000153DaL

Explanation:

We have been given:

15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

This implies that;

1 Da L = 1 x 10²L

So:

1 mL = 1 x 10⁻³L

So 15.3mL will give 15.3 x 10⁻³L

So;

1 x 10²L = 1 DaL

15.3 x 10⁻³L will give $\frac{15.3 x 10^{-3} }{1 x 10^{2} }$ = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL

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A man is walking away from a lamppost with a light source h = 6 m above the ground. the man is m = 2 m tall. how long is the man

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Answer;

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Explanation;

2/x=6/(8+x) cross multiply.

6x=2(8+x)

6x=16+2x

6x-2x=16

4x=16

x=2=16/4

x=4 m. is the length of the man's shadow.

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if the kinetic and potential energy in a system or equal to the potential energy increases what happens as a result

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3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

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3 years ago

if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?

Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

$30 lbf = 30 \times 4.45 = 133.45 N$

now we can use this to find the gravity at this height

$mg = 133.45$

$90\times g = 133.45$

$g = 1.49 m/s^2$

now we know that with height gravity varies as

$g' = g(\frac{R}{R+H})^2$

$1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2$

$h = 1.0 \times 10^7 m$

so above is the height from the surface of earth

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