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3 years ago

Please someone help me plz I don’t understand 1 or 2 plz help me plz plz plz

Physics

1 answer:

Sati [7]3 years ago

7 0

Answer: 1.Magnification and resolution are important because the magnification in larges the image and the resolution makes it move shape and detailed creating the perfect image. 2. i dont know the second one, sorry

Explanation:

You might be interested in

In three to five sentences, identify the acid, base, conjugate acid, and conjugate base. Use patterns in the periodic table to e

Oliga [24]

A conjugate acid is formed from the base by accepting a proton from the acid .

A conjugate base is obtained from the Brownstead - Lowry acid when it looses a proton while the conjugate acid is obtained from the Brownstead - Lowry base when it accepts a proton. In the Brownstead - Lowry sense, acid base reaction involves the loss or gain of a proton.

Consider the hypothetical reaction; AH + :B ⇄ BH + :A. The specie BH is the conjugate acid while the specie :B is the Brownstead - Lowry base . The specie :A is the conjugate base while the specie AH is the Brownstead - Lowry acid.

Learn more about conjugate acid: brainly.com/question/10468518

6 0

2 years ago

a runner covers the last straight stretch of the race in 4 s. durning that time, he speeds up from 5 m/s to 9m/s what is the run

Sergio039 [100]

Answer:

The runner's acceleration was $1\ m/s^2$

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

$v_f=v_o+at$

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:

$\displaystyle a=\frac{9-5}{4}$

$\displaystyle a=\frac{4}{4}=1$

The runner's acceleration was $1\ m/s^2$

5 0

3 years ago

Uranium-238 eventually decays into

GalinKa [24]

Uranium-238 decays<span> by alpha emission </span>into<span> thorium-234, which itself </span>decays<span> by beta emission to protactinium-234, which </span>decays<span> by beta emission to </span>uranium<span>-234, and so on. The various </span>decay<span> products, (sometimes referred to as “progeny” or “daughters”) form a series starting at </span>uranium-238<span>.</span>

4 0

3 years ago

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

3 years ago

Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0

3 years ago