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3 years ago

Which is part of a proper graph?

1 answer:

7 0

The answer is, CC.consistent intervals on each axis. Intervals on each axis should be consistent for easier interpretation of the graph.

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3. A scooter with a mass of 24 kg accelerates at 14 m/s/s when ______ N of force is applied.

wariber [46]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

Force = 24 × 14

We have the final answer as

Hope this helps you

7 0

3 years ago

A circuit carries a current of 0.64 A when there is a resistance of 50.0 Ω. The voltage applied to the circuit, to the nearest w

Natasha2012 [34]

V = I • R

V = 32 volts

3 0

2 years ago

. Egbert is in a rowboat four miles from the nearest point on Egbert a straight shoreline. He wishes to reach a house 12 miles f

swat32

Answer:

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

Explanation:

Given:

Distance from the shore, $s=4\ mi$

distance of house from the shore, $h=12\ mi$

speed of rowing, $v_r=3\ mi.hr^{-1}$

speed of walking, $v_w=4\ mi.hr^{-1}$

<em>For the least amount of time to reach the house one must row at the nearest point on the shore and then walk from there.</em>

<u>Now the time taken to reach the shore:</u>

$t_s=\frac{s}{v_s}$

$t_s=\frac{4}{3}\ mi.hr^{-1}$

<u>Time taken in walking to house from the shore:</u>

$t_h=\frac{h}{v_w}$

$t_h=\frac{12}{4}$

$t_h=3\ mi.hr^{-1}$

<u>Therefore total time taken:</u>

$t=t_h+t_s$

$t=\frac{13}{3} =4.33\ mi.hr^{-1}$

6 0

3 years ago

A ball is dropped from a height h and falls the last half of its distance in 4 seconds. How long does the ball fall? From what h

dlinn [17]

Answer:

How long does the ball fall is t_2 = 13.66 (s).

From what height is the ball originally dropped is h= 913.90 (m).

Explanation:

8 0

3 years ago

If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago