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8 days ago

How are uniform circular motion maps the same as linear motion maps? Check all that apply.

Physics

2 answers:

Ostrovityanka [2.8K]8 days ago

7 0

Answer:

Dots are used to denote time at one-second intervals.

The direction of velocity vectors varies as the object alters its course.

Explanation:

I completed the problem and got it correct! :)

serg [3.2K]8 days ago

3 0

A: Time is depicted by dots spaced at one-second intervals. D: The direction of velocity vectors shifts in tandem with the object's changing direction.

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"a very light ideal spring stretches by 21.0 cm when it is used to hang a 135-n object. what is the weight of a piece of electro

Keith_Richards [2907]

In the first case, once equilibrium is established,
mg = kx
135 = k(0.21) [Converted to SI units]
k = 135/(0.21)

In the second case, similarly when reaching equilibrium,
mg = kx
mg = [(135)(.449)]/(0.21)
mg = weight of the electronic equipment = approximately 289 N.

6 0

27 days ago

Read 2 more answers

A rod 12.0 cm long is uniformly charged and has a total charge of -20.0 µc. determine the magnitude and direction of the electri

Ostrovityanka [2820]

Let Q be the charge, thus Q = -20.0 µC.
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
L = 0.12 m, which represents the length of the rod
To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
E = K·Q/r²
orE = kQ/D(D+L), where k is a constant equal to 8.99 x 109 N m
2/C2.Consequently,[TAG_21]]E=(8.99 x 109 N m2/C2.* (-20.0 µC))/(0.2 m*0.32m)

7 0

13 days ago

Read 2 more answers

Jeff's body contains about 5.46 L of blood that has a density of 1060 kg/m3. Approximately 45.0% (by mass) of the blood is cells

ValentinkaMS [3091]

Answer:

a) Blood mass is $m= 5.7876kg$

b) The count of blood cells is $N_t=1.04*10^{13}$

Explanation:

From the problem statement, we learn that

The blood volume is $V_b = 5.46 \ L = \frac{5.46}{1000} = 0.00546m^3$

The density of blood is $\rho_b = 1060 kg/m^3$

% of blood which consists of cells is = 45.0%

the % of blood that is plasma is = 55.0%

density of blood cells is $\rho_d = 1125kg/m^3$

% of cells that are white is = 1%

% of cells that are red is = 99%

The red blood cell diameter is = $7.5 \mu m = 7.5*10^{-6}m$

The red blood cell radius is $= \frac{7.5*10^{-6}}{2} = 3.75*10^{-6}m$

The mass is generally represented mathematically as

$m = \rho_b * V_b$

Substituting values

$m = 1060 * 0.00546$

$m= 5.7876kg$

Cell mass is $m_c$ = 45% of m

= 0.45 * 5,7876

= 2.60442 kg

The volume of cells is $V_c$ = $\frac{m_c}{\rho_d}$

$= \frac{2.60442}{1125}$

$= 2.315 *10^{-3} m^3$

The white blood cells volume is $V_w$ = 1% of the cells volume

$= \frac{1}{100} * 2.315*10^{-3}$

$= 2.315*10^{-5}m^3$

The volume of a single cell is $V_s = 4 \pi r^3$

$= 4*(3.142) * (3.75*10^{-6})^3$

$= 2.21*10^{-16}m^3$

The red blood cells volume is $V_r = V_c - V_w$

$=2.315*10^{-3} - 2.315*10^{-5}$

$= 2.29*10^{-3}m^3$

The total red blood cell count is $= \frac{V_r}{V_s}$

$= \frac{2.29 *10^{-3}}{2.21*10^{-16}}$

$= 1.037*10^{13}$

The total white blood cell count is $=\frac{V_w}{V_s}$

$= \frac{2.315 * 10^{-5}}{2.21*10^{-16}}$

$= 1.04*10^{11}$

The overall number of blood cells is $N_t= 1.037*10^{13} + 1.04*10^{11}$

$N_t=1.04*10^{13}$

6 0

11 days ago

Read 2 more answers

An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu

Yuliya22 [2968]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>

5 0

1 month ago

A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

kicyunya [2911]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

$U_1=k\dfrac{q_1q_2}{x_1}$

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Consequently, the work done by the electrostatic force on the moving charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Therefore, this concludes the solution.

3 0

20 days ago