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Hunter-Best [27]
3 years ago
13

You hold a spherical salad bowl 50 cm in front of your face with the bottom of the bowl facing you. The salad bowl is made of po

lished metal with a 44 cm radius of curvature.
(a) Where is the image of your 5.0-cm-tall nose located?
(b) What are the image’s size, orientation, and nature (real or virtual)?
Physics
1 answer:
nataly862011 [7]3 years ago
4 0

Answer:

a) q = 39.29 cm ,  b)   h ’= - 3.929 cm  the image is inverted  and REAL

Explanation:

For this exercise we will use the equation of the constructor

          1 / f = 1 / p + 1 / q

where f is the focal length of the salad bowl, p and q are the distance to the object and the image

The metal salad bowl behaves like a mirror, so its focal length is

           f = R / 2

           f = 44/2

           f = 22 cm

a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image

           1 / q = 1 / f  - 1 / p

           1 / q = 1/22 - 1/50

           1 / q = 0.0254

            q = 39.29 cm

b) to calculate the size of the image we use the equation of magnification

           m = h’/ h = - q / p

            h ’= - q / p h

            h ’= - 39.29 / 50 5

            h ’= - 3.929 cm

the negative sign means that the image is inverted

as the rays of light pass through the image this is REAL

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Two astronauts are floating together with zero speed in a gravity-free region of space. The mass of astronaut A is 110 kg and th
Kobotan [32]

Answer:

The recoil speed of Astronaut A is 0.26 m/s.            

Explanation:

Given that,

Mass of astronaut A, m_A=110\ kg

Mass of astronaut B, m_B=74\ kg

Astronaut A pushes B away, with B attaining a final speed of 0.4, v_B=0.4\ m/s

We need to find the recoil speed of astronaut A. The momentum remains conserved here. Using the law of conservation of linear momentum as :

m_Av_A=m_Bv_B\\\\v_A=\dfrac{m_Bv_B}{m_A}\\\\v_A=\dfrac{74\times 0.4}{110}\\\\v_A=0.26\ m/s

So, the recoil speed of Astronaut A is 0.26 m/s.                                

4 0
2 years ago
A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area
Airida [17]
For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
p_1 = p_2
which becomes
\frac{F_1}{A_1}= \frac{F_2}{A_2}
where
F_1=3600 N is the force on the first piston
A_1=0.030 m^2 is the area of the first piston
F_2 is the force on the second piston
A_2=0.015 m^2 is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
F_2=F_1  \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N
7 0
3 years ago
Read 2 more answers
A series circuit consists of a 100-ω resistor, a 10.0-μf capacitor, and a 0.350-h inductor. the circuit is connected to a 120-v
Tpy6a [65]
Current will be I=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{\sqrt{ R^{2}+(X_{C}-X_{L})^{2}}}\\where~X_{c}=\dfrac{1}{j.\omega .C}~and~X_{L}=j.\omega.L~where~\omega=2.\pi f~and~f=60Hz
now just pluf in the values and Voila..
7 0
3 years ago
Two small children decide it would be fun to toss a couple of large cats at each other. Cat A (7kg) is thrown at 7m/s and cat B
Alex777 [14]

Answer:

V=4.7m/s

Explanations:

Let Ma mass of cat A=7kg

Va velocity of cat A=7m/s

Mb mass of cat b=6.1kg

VB velocity of cat b=2m/s

From conservation of linear momentum

MaVa+MbVb=(Ma+Mb)V

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3 0
3 years ago
011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

8 0
3 years ago
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