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3 years ago

How would I label the y axis and fill it out and answer questions at the bottom I need answers please and thanks

Physics

1 answer:

yKpoI14uk [10]3 years ago

7 0

you must write the temprature of the copper sulfate solution on the on the y axis

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When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch

vampirchik [111]

Your Answer Will Be C

8 0

3 years ago

The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

3 years ago

Five wheels are connected as shown in the figure. Find the velocity of the block “Q”, if it is known that: RA= 5 [m], RB= 10 [m]

Tanzania [10]

Answer:

-5 m/s

Explanation:

The linear velocity of B is equal and opposite the linear velocity of E.

vB = -vE

vB = -ωE rE

10 m/s = -ωE (12 m)

ωE = -0.833 rad/s

The angular velocity of E is the same as the angular velocity of D.

ωE = ωD

ωD = -0.833 rad/s

The linear velocity of Q is the same as the linear velocity of D.

vQ = vD

vQ = ωD rD

vQ = (-0.833 rad/s) (6 m)

vQ = -5 m/s

6 0

3 years ago

A 10 kg cart is rolling across a parking lot with a velocity of .5 m/s. What is its momentum?

-Dominant- [34]

Hello!

$\large\boxed{p = 5 \text{ } kgm/s}$

Use the equation for momentum:

$p = mv$

Plug in the given mass and velocity into the equation:

$p = 10 * 0.5$

$p = 5 kgm/ s$

4 0

3 years ago

A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal

slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

$d_0=5m=500cm$

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

$\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm$

By the formula of magnification

$\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm$

The height of the image formed is 18.89cm.

5 0

3 years ago