a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power

Answer:

Gold = 19.3g/cm³

Explanation:

Given the following data;

Mass = 38.6g

Volume = 2cm³

To find the density;

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

[tex]Density = \frac{mass}{volume}[/tex]

Substituting into the equation, we have;

[tex]Density = \frac{38.6}{2}[/tex]

Density = 19.3g/cm³

Therefore, from the table the material is Gold.

Gold is a chemical element and it is the element 79 on the periodic table. Thus, it has an atomic number of 79. The chemical symbol for Gold is Au, it is chemically classified as a transition metal and as a solid at room temperature.

Generally, the chemical element Gold is known to have the following physical properties;

I. Malleable.

II. Ductile.

III. A good conductor of electricity and heat.

Also, it is a non-toxic chemical element with a beautiful lustrous sheen.

Answer:

speed = distance/time

Explanation:

distance -> s

speed -> v

time -> t

The calculation of speed using time and distance shows that speed is equal to distance/time.

Speed refers to the rate of movement of a vehicle or person.

Thus, the speed can be computed by dividing the total distance by the time consumed.

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Answer:

Approximately [tex]3.99\times 10^{4}\; \rm J[/tex] (assuming that the melting point of ice is [tex]0\; \rm ^\circ C[/tex].)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

[tex]\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}[/tex]

The energy required comes in three parts:

The following equation gives the amount of energy [tex]Q[/tex] required to raise the temperature of a sample of mass [tex]m[/tex] and specific heat capacity [tex]c[/tex] by [tex]\Delta T[/tex]:

[tex]Q = c \cdot m \cdot \Delta T[/tex],

where

For the first part of energy input, [tex]c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}[/tex].

Similarly, for the third part of energy input, [tex]c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}[/tex] whereas [tex]m = 0.100\; \rm kg[/tex]. Calculate the change in the temperature:

[tex]\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}[/tex].

Calculate the energy required to achieve that temperature change:

[tex]\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}[/tex].

The second part of energy input requires a different equation. The energy [tex]Q[/tex] required to melt a sample of mass [tex]m[/tex] and latent heat of fusion [tex]L_\text{f}[/tex] is:

[tex]Q = m \cdot L_\text{f}[/tex].

Apply this equation to find the size of the second part of energy input:

[tex]\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}[/tex].

Find the sum of these three parts of energy:

[tex]\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}[/tex].

Answer: 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final  velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE =  ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

Answer:

Alfred Wegener

Explanation:

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm[/tex]

Magnification,

[tex]m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25[/tex]

Hence, the magnification of the image is 0.25.

Answer: Biosensors are modern electronic devices which are made up of biological recognition system and a transducer, for processing of signals and to quantify a particular analyte.

Explanation:

In modern medicine, a Biosensors are used to replace some clinical laboratory investigations of biological fluids with the advantages of being easy to use, rapid and robust as well as offering multianalyte testing capability. They are classified based on their biological recognition elements which includes:

--> enzymatic biosensors

--> immuno biosensors

--> DNA and whole cell biosensors.

They can also be classified according to their signal transduction methods. These includes:

--> electrochemical Biosensors,

--> optical biosensors,

--> thermal and mass-based biosensors;

Some of the HEALTH APPLICATIONS of a modern biosensor includes:

--> Monitoring of glucose levels in diabetic patients which makes use of enzyme based sensors to monitor concentration of glucose in patients.

--> Detection of general metabolic status of bacteria, fungi, yeast, animal or plant cells.

--> Monitoring of cholesterol levels to prevent cardiovascular diseases. Cardiac troponin and. C - Reactive Proteins are biomarkers that are easily detected using biosensors.

--> They are used for Cancer clinical testings

Answer:

the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Explanation:

The computation of the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is as follows:

The ball momentum is

[tex]\Delta Q = mv\\\\\Delta Q = 6 \times 1^-2 \times 60\\\\\Delta Q = 3.6 kg \times m/s[/tex]

As there is a similar momentum

So,

[tex]\Delta Q = MV\\\\3.6 = 10^-2V\\\\V = 360 m/s[/tex]

Hence, the speed that have 0.0100 kilogram bullet with the similar momentum magnitude is 360 m/s

Answer:

100 m

Explanation:

Answer:

because

Explanation:

because we need eye to see if you don't have eyes how you regnise some one

if you have ears how can you know who's voice is who's

if don't have fingers how can you know what your are holding this is why they are called physical sensors

Answer:

because  they stimulate our brain and the sensor of human body detect stimuli.

Explanation:

Answer:

m = 4.4 × 10³ kg

Explanation:

Given that:

The total yearly energy is 4.0 × 10²⁰ J

The amount of mass that provides this energy can be determined by using the formula:

E = mc²

where;

c = speed of light in free space = (3 × 10⁸)

4.0 × 10²⁰ = m × (3 × 10⁸)²

[tex]m = \dfrac{4.0 \times 10^{20} }{(3\times 10^8)^2}[/tex]

m = 4.4 × 10³ kg

Answer:

16m/s

Explanation:

[tex]v_{f}=v_{i}+at[/tex]

[tex]v_{f}=0+2\cdot8[/tex]

[tex]v_{f}=16\ \frac{m}{s}[/tex]

Therefore,  the speed after 8 seconds is 16m/s

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

Answer:

Density = 5 g/m³

Explanation:

Given:

Mass of block = 20 gm

Volume of block = 4 m³

Find:

Density

Computation:

Density = Mass / Volume

Density = 20 / 4

Density = 5 g/m³

5 kg/m3

Explanation:

The formula to find density is (d) = mass (m) ÷ volume (v)

Answer:

58.8

Explanation:

we should apply formula

m*g*h

3*9.81*2

Answer:

D. Nuclear fission is used to power submarines and aircraft carriers.

Explanation:

A and B describe nuclear fusion. C doesn't apply to nuclear fusion or nuclear fission.

Solar Energy is the answer to the question tell me if i`m wrong

Answer: its linear momentum is 1.78 × 10²⁹ kg.m/s

Explanation:

Given that;

mass of Earth m =  5.972 x 10²⁴ kg

radius r = 1.496 x 10¹¹ m

period t = 3.15 x 10⁷ s

now we know that Earth rotates in a circular path so the distance travelled per rotation is;

d = 2πr we substitute

d = 2π × 1.496 x 10¹¹ m

= 9.4 × 10¹¹ m

Now formula for speed v is;

v = d/t

we substitute

v = 9.4 × 10¹¹ m / 3.15 x 10⁷ s

v = 2.98 × 10⁴ m/s

now we determine the linear momentum p

linear momentum p = mv

we substitute

p = (5.972 x 10²⁴ kg) × (2.98 × 10⁴ m/s)

p = 1.78 × 10²⁹ kg.m/s

Therefore its linear momentum is 1.78 × 10²⁹ kg.m/s

The linear momentum of earth is 17.8 * 10²⁸ kgm/s

The orbital radius (r) of earth is 1.496 x 10¹¹ m. Hence the distance covered by one rotation is:

Distance = 2πr = 2π(1.496 x 10¹¹) m

The period of one rotation is 3.15 x 10⁷ s.

The velocity of earth (v) = distance/time = 2π(1.496 x 10¹¹) m/ 3.15 x 10⁷ s = 298840 m/s

Linear momentum = mass * velocity = 5.972 x 10²⁴ kg * 298840 m/s = 17.8 * 10²⁸ kgm/s

Therefore the linear momentum of earth is 17.8 * 10²⁸ kgm/s

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Answer:

The value is  [tex]u = 72.69 \ m/s[/tex]

Explanation:

From the question we are told that

The amount of force a square meter of an aircraft wing should produce is [tex]F = 1000 \ N[/tex]

 The air speed relative to the bottom of the wing is  [tex]v = 61.1 \ m/s[/tex]

   The air level density of air is  [tex]\rho_s = 1.29\ kg/m^3[/tex]

Gnerally this  force per square meter  of an aircraft wing is mathematically represented as

              [tex]F = \frac{1}{2} * \rho_s * A * [ u^2 - v^2 ][/tex]

Here u is the speed air need to go over the top surface to create the ideal lift

          A  is the area of a square meter i.e   [tex]A = 1 \ m^2[/tex]

So              

           [tex]1000 = \frac{1}{2} * 1.29 * 1 * [ u^2 - 61.1 ^2 ][/tex]

=>         [tex]u = 72.69 \ m/s[/tex]