Nate throws a ball straight up to Kayla, who is standing on a balcony 3.8 m above Nate. When she catches it, the ball is still moving upward at a speed of 2.8 m/s. Required:With what initial speed did Nate throw the ball?

Answer:

its right the way it is

Explanation:

if there is a multiple choice then pick 20 v

Answer:

The mass in grams of a red blood cell is about 7.85 ×  10⁻¹¹ grams

Explanation:

To find the mass in grams of a red blood cell,

From,

[tex]Density = \frac{Mass}{Volume}[/tex]

Then,

[tex]Mass = Density \times Volume[/tex]

From the question,

Density of a red blood cell is similar to that of water

Density of water = 1 g/mL = 1 g/ cm³

Then, Density of a red blood cell = 1 g/cm³

Now, we will find the volume a red blood cell.

From the question,  

Since the shape is like that of a thick disc, we can determine the volume by using the formula for volume of a cylinder.

Hence,

Volume of a red blood cell = [tex]\pi r^{2}h[/tex]

Where [tex]\pi[/tex] Is a constant (Take [tex]\pi[/tex] = 3.14)

[tex]r[/tex] is the radius

and [tex]h[/tex] is the thickness

Diameter of a red blood cell = 10 micrometers

Then, radius of a red blood cell = 10/2 micrometers = 5 micrometers

[tex]r[/tex] = 5 micrometers = 5 × 10⁻⁶ meters

and [tex]h[/tex] = 1 micrometer = 1 × 10⁻⁶ meters

Hence,

Volume of a red blood cell = 3.14 × (5 × 10⁻⁶)² × 1 × 10⁻⁶

∴ Volume of a red blood cell = 7.85 × 10⁻¹⁷ cubic meter (m³)

Convert this to cubic centimeter

(NOTE: 1 cubic meter = 1000000 cubic centimeter)

Hence,

Volume of a red blood cell = 7.85 ×  10⁻¹¹ cubic centimeter (cm³)

Now, for the mass

[tex]Mass = Density \times Volume[/tex]

Density of a red blood cell = 1 g/cm³

Volume of a red blood cell = 7.85 ×  10⁻¹¹ cubic centimeter (cm³)

Then,

Mass = 1 g/cm³ ×  7.85 ×  10⁻¹¹ cm³

Mass = 7.85 ×  10⁻¹¹ g

Hence, the mass in grams of a red blood cell is about 7.85 ×  10⁻¹¹ grams

Answer:

Choice A. Without energy storage, the total work output of a machine will not exceed the total work input.

Explanation:

Work is the product of force and the distance travelled in the direction of that force.

Indeed, simple machines can alter the size or direction of a force. However, their work output can't exceed the work input to that machine (unless the machine stores some energy- by bending, for example.) Because of frictions and other forms of energy loss, the useful work output of a real machine can even be smaller than the work input to that machine.

The force output of some machines can certainly exceed their force input. However, at the same time, the input part of those machine needs to move a  distance that is (more than) proportionally longer compared to the output part of the machine. As a result, the useful work output would be equal to or even smaller than the total work input.

A fixed pulley is an example of a machine that applies a force in a direction different from the direction of the force applied to the machine. Pulling the string down will lift up the weight on the other side. A seesaw (a first-class lever) is another real-world example of this kind of machines.

Tongs are examples of machines where: the distance traveled by the output exceeds that of the input. However, that comes at a price: delivering the same amount of force with a pair of tongs requires more force than without a pair of tongs. That corresponds to choice D: the work output of some machines can be less than the force input to these machines.

At the same time, some extra force input might be required to overcome the friction between parts of the machine. That's another reason why the machine applies less force than the input even if it was designed to apply the same amount of force.

Answer:

A.) Do a greater amount of work than the amount of work done on the machine.

Explanation:

I took the quiz on Edge!

Hope this helped!

Answer:

2.15m/26.1s is = average speed

0.08 m/s

Answer:

Thus simply tells us that The nut has no net charge and so therefore, There will be a negative charge on the left side, and an equal positive charge on the right side

This basically means that the nut has no net charge and thus will have a negative charge on the left side and an equal positive charge on the right side.

When an object has more protons than electrons, its net charge is positive.

If there are more electrons than protons in an object, the net charge is negative. If the object has an equal number of protons and electrons, it is electrically neutral.

The total charge on a body is the algebraic sum of all the charges on it. Every atom is electrically neutral because it contains the same number of electrons as protons.

This basically means that the nut has no net charge and thus has a negative charge on one side and an equal positive charge on the other.

Thus, this can be implied as per the given scenario.

For more details regarding net charge, visit:

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Answer:

a) 28877 AU

b) 1.795 sec

Explanation:

Time it took the radio signal to reach earth = 4 hours = 4 x 3600 = 14400 s

speed of radio signal = speed of vacuum speed of light  = 3 x 10^8 m/s

Distance traveled by this signal = speed x time

distance = 3 x 10^8 x 14400 = 4.32 x 10^12 m

One Astronomic unit AU is the distance between the Earth and the sun and it is approximately equal to 1.496 x 10^8 m

The distance traveled by the signal in AU = (4.32 x 10^12)/(1.496 x 10^8) = 28877 AU

b) The minimum separation between Earth and Mars = 0.36 AU

This distance = 0.36 x 1.496 x 10^8 = 538560000 m

Time that will be taken for a radio signal to travel this distance = distance/speed

==> 538560000/(3 x 10^8) = 1.795 sec

Answer:

Velocity and Time

Explanation:

I googled it-

Answer:

h = 31.9 m

Explanation:

Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value:

t = 5.1 s /2 = 2.55 s

Now, we use 1st equation of motion between the time of throwing and the time of reaching maximum height:

Vf = Vi + gt

where,

Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point)

Vi = Initial Velocity = ?

g = - 9.8 m/s² (negative sign for upward motion)

Therefore,

0 m/s = Vi + (-9.8 m/s²)(2.55 s)

Vi = 24.99 m/s

Now, we use second equation of motion for height (h):

h = Vi t + (0.5)gt²

h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)²

h = 63.7 m - 31.8 m

h = 31.9 m

Answer:

[tex](a) 189.23 N[/tex], [tex](b) 47.31 N[/tex] and [tex](c) 141.92 N[/tex].

Explanation:

Three balls are shown in figure having charge [tex]q=1.45 \mu C[/tex]. The middle ball, [tex]B[/tex], is positively charged having charge [tex]+q[/tex], and the remaining two outside balls, [tex]A[/tex] and [tex]C[/tex], are negatively charged having charged [tex]-q[/tex] as shown.

[tex]AC=20 cm[/tex] and [tex]AB=BC=10[/tex] cm as B is the mid-point of AC.

Let [tex]d_1=AC=20\times 10^{-3}m[/tex] and [tex]d_2=AB=BC=10\times 10^{-3}m[/tex]

From Coulomb's law, the magnitude of the force, [tex]F[/tex], between two point charges having magnitudes [tex]q_1 \& q_2[/tex], separated by distance, [tex]d[/tex], is

[tex]F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)[/tex]

where, [tex]\epsilon_0[/tex] is the permittivity of free space and

[tex]\frac {1}{4\pi\epsilon_0}=9\times 10^9[/tex] in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

[tex]F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}[/tex]

[tex]\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}[/tex]

[tex]\Rightarrow F_{AB}=189.23 N[/tex]

(a) The magnitude of the repulsive force between balls A and C is

[tex]F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}[/tex]

[tex]\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}[/tex]

[tex]\Rightarrow F_{AC}=47.31 N[/tex]

(c) The magnitude of the net force, [tex]F_{net}[/tex], on the outside of the ball is,

[tex]F_{net}=189.23-47.31 N[/tex]

[tex]\Rightarrow F_{net}=141.92 N[/tex]

Answer:

i try B

Explanation:

Physical properties of matter tells us what a substance is or what a substance can do when it is not undergoing a chemical change.

These properties are observable with our senses or instruments or some pieces of apparatus.

Some of these properties are :

Color

Odor

hardness

texture

boiling point

Melting point

density

viscosity

They differ from chemical properties which are shown by the reaction of a substance with another.

Answer:

-10.267

Explanation:

Initial speed > 35 miles per hour

Final speed > 0 miles per hour

Time > 5 seconds

you have to divide the velocity by the time which should be 35 / 5 = 7

so the answer should be 7 but I'm not sure

Answer:

He examined covered and uncovered meat to determine that maggots came from eggs.

In other other words A

Answer:

[tex]a=-1.5\ m/s^2[/tex]

Explanation:

Given that,

Initial velocity, u = 30 m/s

Final velocity, v = 0

Time, t = 20 s

We need to find the acceleration of the car. The rate of change of velocity is equal to acceleration. So,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{-30}{20}\\\\a=-1.5\ m/s^2[/tex]

So, the magnitude of the acceleration of the car is [tex]1.5\ m/s^2[/tex] and it is decelerating.

Complete Question

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 55.0 min at 60.0 km/h, 18.0 min at 80.0 km/h, and 60.0 min at 60.0 km/h and spends

25.0 min eating lunch and buying gas

What is the total distance traveled over the entire trip (in km)

Answer:

The value is [tex]D = 139.02 \ km[/tex]

Explanation:

From the question we are told that

For [tex] t_1 = 55 min = \frac{55}{60} = 0.917 \ h[/tex] the speed is [tex]v_1 = 60 \ km/h[/tex]

For [tex] t_2 = 18 min = \frac{18}{60} = 0.3 \ h[/tex] the speed is [tex]v_2 = 80 \ km/h[/tex]

For [tex] t_3 = 60 min = \frac{60}{60} = 1 \ h[/tex] the speed is [tex]v_3 = 60\ km/h[/tex]

The time taken to have lunch is [tex]t _l = 25 \ min = \frac{25}{60} = 0.42 \ h[/tex]

Generally the total distance traveled over the entire trip (in km) is mathematically represented as

[tex]D = t_1 * v_1 + t_2 * v_2 + t_3 * v_3[/tex]

=>   [tex]D = 0.917  *  60 +  0.3 * 80 +  1  * 60[/tex]

=>   [tex]D = 55.02  + 24 +   60[/tex]

=>   [tex]D = 139.02 \ km[/tex]

Answer:53.03

Explanation:

Answer:

A. The potential difference is given as

V =q/C=0.146uC/240pF

608.33 V

B) The area of each plate is

A =Cd/eo =9.9525*10^-3m ²

C) the electric field magnitude between the plates=V/d

608.33/0.367mm

=1.657x 10^6 N/C

D) The surface-charge density on each plate=q/A

0.146uC/9.9525 x 10^-3m ²

=1.466 x 10^-5 C/m²

Answer:

[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]

Explanation:

The options are not well presented; However, the questions can still be solved

Given

[tex]Initial\ Velocity= V_{ox}[/tex]

[tex]Initial\ Displacement = X_{o}[/tex]

[tex]Time = t[/tex]

Required

Determine the final displacement

This question will be answered using the following equation of motion

[tex]v^2 = u^2 + 2as[/tex]

Where s represent the total distance

s is the distance between the initial and final displacements and is calculated as thus;

[tex]s = x - o_x[/tex]

Where x represents the final displacement

Substitute [tex]V_{ox}[/tex] for [tex]u[/tex] ---- The initial velocity

[tex]v^2 = v_{ox}^2 + 2as[/tex]

Substitute [tex]x - o_x[/tex] for s

[tex]v^2 = v_{ox}^2 + 2a(x - o_x)[/tex]

Hence, the above equation can be used to determine the final displacement by solving for x

Answer:

The equation which describes conservation of charge is [tex]Q_{initial} - Q_{final } = 0[/tex]

Explanation:

The law of conservation charge states that for an isolated system that sum of initial charges is equal to sum of final charges, that is the total charge is conserved.

let the sum of initial charges = [tex]Q_{initial}[/tex]

let the sum of the final charges = [tex]Q_{final}[/tex]

[tex]Q_{initial } = Q_{final}\\\\Q_{initial } - Q_{final} = 0[/tex]

Therefore, the equation which describes conservation of charge is [tex]Q_{initial} - Q_{final } = 0[/tex]

Answer:

[tex]Q_{initial}-Q_{final}=0[/tex]

Explanation:

According to the conservation of charge "Charge can neither be created nor destoryed". This line means the net amount of charges remain conserved, no matter what happens.

Equation will be of the form:

Initial Charge =Final Charge

It can also be written as:

Initial Charge- Final Charge=0

Let Q be the charge:

[tex]Q_{initial}-Q_{final}=0[/tex]

The above equation describes conservation of charge.

Answer:

I think its n u c l e a r e n e r g y.

Answer:

Definition - The friction that occurs when an object rolls across a rolling friction is easier to overcome than sliding friction.

Ex - Anything with wheels (cars, bicycle,etc) or ball rolling.

Explanation:

I think this answer will help .... if so please follow me.

Answer: Hypothesis

Explanation:

For each experiment involving nanotechnology, Gerry  and Lena will test a hypothesis.

Gerry and Lana will use the Scientific method for their nanotechnology research and a key stage in the scientific method is to come up with a Hypothesis.

A hypothesis gives a researcher direction because it is a theory that they formulate that is meant to explain the phenomenon that they are researching.

They will then test this theory to either disprove or approve it and in so doing will be able to come up with conclusions to the research. Gerry and Lana will therefore have to test a hypothesis for each experiment in order to continue with their research.

Answer:

Explanation:

For two vectors to be orthogonal (perpendicular) the product of both vectors must be zero. Given the vectors D=(2.0i-4.0j+k)N and G=(3.0i+4.0j+10k)N, to show that they are orthogonal, we will take their dot product as shown;

D.G = (2.0i-4.0j+k).(3.0i+4.0j+10k)

Note that i.1 = j.j = k.k = 1 and dot product of different component is zero.

D.G = 2.0(3.0) (i.i) + (-4.0)(4.0)j.j + 1(10)k.k

D.G = 6.0(1) -(16)(1)+10(1)

D.G = 6-16+10

D.G = -10+10

D.G = 0

Since the dot product of the two vectors is zero, this shows that force vector D is orthogonal to force vector G.

Answer:

a)   t = 57.14 s

b)    x = 27711.4 m

Explanation:

This is a missile throwing exercise

a) They ask us for the time to the maximum reach, this corresponds to when it reaches the ground y = 0, let's use

          y =  [tex]v_{oy}[/tex] t - ½ g t²

Let's use trigonometry to find the vertical initial velocity

         sin θ = v_{oy} / v₀

         v_{oy} = v₀ sin θ

we substitute

         y = v₀ sin θ t - ½ g t²

since the height is zero

          0 = t (v₀ sin θ - ½ g t)

This equation has two solutions

*  t = 0 which corresponds to the moment of launch

*

         v₀ sin 30 - ½ g t = 0

        t = v₀ sin 30    2/g

let's calculate

         t = 560 sin 30  2 / 9.8

         t = 57.14 s

b) with this time we can calculate the distance traveled

          x = v₀ₓ t

let's use trigonometry for velocity

          cos θ = v₀ₓ / v₀

          v₀ₓ = v₀ cos 30

we substitute

         x = v₀ cos 30     t

let's calculate

         x = 560 cos 30 57.14

         x = 27711.4 m

Answer:

0.59 seconds

Explanation:

Answer:

The magnitude of the velocity is approximately 11.45 units

Explanation:

Since they give us the x and y components of the vector velocity, we can find its magnitude via the Pythagorean theorem:

[tex]|v|=\sqrt{9.5^2+(-6.4)^2} \approx 11.45[/tex]

Answer:

a) The acceleration of the sprinter is 6.521 meters per square second, b) The acceleration of 84512.16 kilometers per square hour.

Explanation:

Note: Statement is incomplete, complete description of the problem is: A sprinter accelerates from rest to [tex]9.00\,\frac{m}{s}[/tex] in 1.38 seconds. What is his acceleration in a. [tex]\frac{m}{s^{2}}[/tex] and b. [tex]\frac{km}{h^{2}}[/tex]?

a) We assume that sprinter accelerates uniformly, so that acceleration ([tex]a[/tex]), measured in meters per square second, can be obtained from this kinematic expression as function of initial and final velocities ([tex]v_{o}[/tex], [tex]v[/tex]), measured in meters per second, and time ([tex]t[/tex]), measured in seconds, as well.

[tex]v = v_{o} + a\cdot t[/tex]

[tex]a = \frac{v-v_{o}}{t}[/tex]

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 9\,\frac{m}{s}[/tex] and [tex]t = 1.38\,s[/tex], the acceleration experimented by the sprinter is:

[tex]a = \frac{9\,\frac{m}{s}-0\,\frac{m}{s} }{1.38\,s}[/tex]

[tex]a = 6.521\,\frac{m}{s^{2}}[/tex]

The acceleration of the sprinter is 6.521 meters per square second.

b) If we know that a hour is equivalent to 3600 seconds and a kilometer is equivalent to 1000 meters, the result of the previous item is converted by using two consecutive unit conversions:

[tex]a = \left(6.521\,\frac{m}{s^{2}} \right)\cdot \left(\frac{1}{1000}\,\frac{km}{m} \right)\cdot \left(3600^{2}\,\frac{s^{2}}{h^{2}} \right)[/tex]

[tex]a = 84512.16\,\frac{km}{h^{2}}[/tex]

The acceleration of 84512.16 kilometers per square hour.

The magnetic field is just a unit vector that explains the electromagnetic influence on electric currents, current flow, and magnetic fluids.

Current density [tex]J = ar^2[/tex]

value of the constant is=  [tex]28.5 \ \frac{A}{mm^4}[/tex]

radius = 5.20 mm

magnetic permeability [tex]\mu = 4\pi \times 10^7 \ \frac{N}{A^2}[/tex]

calculating the area element for the straight circular conduction rod
:

[tex]d_A=2\pi r dr[/tex]

Calculating the current, which is carried in the rod

[tex]dI = \int dA \vec{J}[/tex]

Calculating the above equation with the limit value that is 0 to r.

[tex]I^{1}=\int_{0}^{r} ar^2 \times 2 \pi \cdot r d_r[/tex]

    [tex]=2\pi a\int_{0}^{r} r^3 d_r \\\\=\frac{\pi ar^4}{2}[/tex]

The calculated current value which is carried by the rod is [tex]\boxed{\frac{\pi ar^4}{2}}[/tex]

In option (a)

calculating the magnitude of the magnetic field at the point [tex]r_1= \frac{R}{2}[/tex]

[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r}\\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}r^4)\\\\\to B \times 2\pi = \mu_{0} (\frac{\pi a}{2}r^3)\\\\\to B= \frac{\mu_{0}}{4}r^3 a\\\\[/tex]

Substituting the above value:[tex]\frac{R}{2} \ for \ r[/tex]

[tex]B= \frac{\mu_{0}}{4}(\frac{R}{2})^3 a[/tex]

[tex]B= \frac{4 \pi \times 10^{-7}}{4}(\frac{5.2 \times 10^{-3}}{2})^3 \frac{28.5}{10^{-12}}[/tex]

   [tex]= \frac{4 \pi \times 10^{-7}}{4} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{1} \times \frac{140.608 \times 10^{-9}}{8} \times \frac{28.5}{10^{-12}}\\\\=\frac{1572.87624 \times 10^{-16}}{ 10^{-12}}\\\\=0.157 \ \ T[/tex]

Thus, the magnitude of the magnetic field at the point [tex]r_1 =\frac{R}{2}[/tex] is [tex]\boxed{0.157 \ T}[/tex]

In option (b)
In this, we calculate the magnitude of the magnetic field at the point [tex]r_2= 2R[/tex]

[tex]\to B=\frac{\mu_{0} I^{1}}{2 \pi r} \\\\\to B \times 2\pi r= \mu_{0} (\frac{\pi a}{2}R^4)\\\\\to B \times 2\pi 2R = \mu_{0} (\frac{\pi a}{2}R^3)\\\\\to B= \frac{\mu_{0}}{8}R^3 a\\\\[/tex]

Substituting the values  

[tex]B= \frac{4 \pi \times 10^{-7}}{8}(5.2 \times 10^{-3})^3(\frac{28.5}{10^{-12}})[/tex]

   [tex]= \frac{4 \times 3.14 \times 10^{-7}}{8} \times (5.2)^3 \times (10^{-3})^3 \times\frac{28.5}{10^{-12}}\\\\= \frac{ 3.14 \times 10^{-7}}{2} \times 140.608 \times 10^{-9} \times\frac{28.5}{10^{-12}}\\\\= 6291.50495 \times 10^{-4}\\\\= 0.629 \ \ or \ \ 0.63\\[/tex]

Thus, the magnitude of the magnetic field at the point [tex]r_2 = 2R[/tex] is [tex]\boxed{0.63 \ \ T}[/tex]

Find out more about Magnetic fields here:

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Answer:

a) 5000 g

b) 5250 g

c) 5150 g

Explanation:

For easier calculations, the formulas will be converted from g/cm³ to kg/m³, and then back when we're done.

Density of pure water is 1 g/cm³

1 g/cm³ = 1 * 0.001/0.000001

1 g/cm³ = 1000 kg/m³, and thus,

Density of pure blood

1.05 g/cm³ = 1050 kg/m³

Density of seawater

1.03 g/cm³ = 1030 kg/m³

Recall that, Density = mass / volume, and as such, mass = density * volume.

Converting our volume from L to m³

1 m³ = 1000 L, and as such

1 L = 0.001 m³

5 L = 0.005 m³

Mass of pure water = 1000 * 0.005

Mass of pure water = 5 kg

Mass of pure blood = 1050 * 0.005

Mass of pure blood = 5.25 kg

Mass of seawater = 1030 * 0.005

Mass of seawater = 5.15 kg

Converting these masses back to g, we have

Mass of pure water = 5 kg * 1000 g

Mass of pure water = 5000 g

Mass of pure blood = 5.25 kg * 1000 g

Mass of pure blood = 5250 g

Mass of seawater = 5.15 kg * 1000 g

Mass of seawater = 5150 g