What kind of force is the force of gravity?

According to the solubility curve, KCl is soluble in water at a rate of about 40 grammes per 100 g. Given that we have 100 mL, or 100 grammes, of water, the mass of KCl that will dissolve is similarly 40 grammes.

[tex]Ce_2(SO_4)_3[/tex] is soluble in water at a rate of around 30 grammes per 100 grammes at 100 °C. [tex]Ce_2(SO_4)_3[/tex] will therefore dissolve in 100 mL of water at a mass of 30 grammes.

[tex]NH_4Cl[/tex] dissolves in water at a rate of around 90 grammes per 100 grammes at 90 °C. The mass of [tex]NH_4Cl[/tex] that will dissolve in 100 mL of water is similarly 90 grammes.

We must ascertain the solubilities of the abovementioned chemicals at 15°C in order to ascertain which substance is most soluble at that temperature.

[tex]NaNO_3[/tex] has the highest solubility at 15°C, as seen by the solubility curve. At 15°C, [tex]NaNO_3[/tex] dissolves in water at a rate of about 80 grammes per 100 g.

Therefore, at 15°C, NaNO3 is the most soluble substance among the three given substances.

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Answer:

[tex]W=70 * 5cos20 = 328.89 J[/tex]

[tex]W_n = 0[/tex]

[tex]W_g=0[/tex]

[tex]W_f= -184.59J[/tex]

Work done is 0

Explanation:

From the question we are told that

Weight of block =15.0kg

Force acting on the block = 70.0N

At an angle of 20 degree

Displacement of block is 5m

Coefficient of kinetic friction 0.3

b) Generally work done by force is give by [tex]W=fdcos \theta[/tex]

therefore

      [tex]W=70 * 5cos20 = 328.89 J[/tex]

c) there is no work done by the normal force in this scenario because

normal force in this case is perpendicular to the displacement of the motion

       [tex]W_n = 0[/tex]

d) The displacement in the vertical direction is 0

Therefore the gravitational work done is 0  [tex]W_g=0[/tex]

e)Generally in finding work done by friction we first find frictional force

Mathematically the equation for frictional force is given [tex]f = \alpha N[/tex]

Given that

       [tex]N=mg-Fsin20[/tex]

       [tex]N= 15.0*9.8 - 70 sin20[/tex]

       [tex]N=123 N[/tex]

       [tex]f=0.3* 123.06 = 36.92N[/tex]

Mathematically solving to get work done by frictional force [tex]W_f[/tex]

        [tex]W_f= -fd\\W_f = -36.92 * 5[/tex]

         [tex]W_f= -184.59J[/tex]

the frictional force work done is  [tex]W_f= -184.59J[/tex]

Answer:

approx 6 × 10^24 kg is earth's mass

Answer:

40m/s

Explanation:

144km/h

1km=1000m

1hr=3600secs

144×1000/3600=

40m/s

Answer:

[tex]N_s : N_p = 20 : 1[/tex]

Explanation:

From the question we are told that

    The primary voltage is  [tex]V_p = 120 \ V[/tex]

     The secondary voltage is  [tex]V_s = 6 \ V[/tex]    

Generally from the transformer equation we have that

        [tex]\frac{V_p}{V_s} = \frac{N_p}{N_s}[/tex]

So

       [tex]\frac{120}{6} = \frac{N_p}{N_s}[/tex]

=>      [tex]\frac{N_p}{N_s} = 20[/tex]

Therefore the ratio of the number of turns in the secondary to the number of turns in the primary is  

       [tex]N_s : N_p = 20 : 1[/tex]

Answer:

Alpha particles carry a positive charge, beta particles carry a negative charge, and gamma rays are neutral. An alpha particle is made up of two protons and two neutrons bound together. Beta particles are high energy electrons. Gamma rays are waves of electromagnetic energy, or photons.

Answer:

(a)  the runner's kinetic energy at the given instant is 308 J

(b)  the kinetic energy increased by a factor of 4.

Explanation:

Given;

mass of the runner, m = 64.1 kg

speed of the runner, u = 3.10 m/s

(a) the kinetic energy of the runner at this instant is calculated as;

[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 64.1 \times 3.1^2\\\\K.E_i = 308 \ J[/tex]

(b) when the runner doubles his speed, his final kinetic energy is calculated as;

[tex]K.E_f = \frac{1}{2} mu_f^2\\\\K.E_f = \frac{1}{2} m(2u)^2\\\\K.E_f = \frac{1}{2} \times 64.1 \ \times (2\times 3.1)^2\\\\K.E_f = 1232 \ J[/tex]

the change in the kinetic energy is calculated as;

[tex]\frac{K.E_f}{K.E_i} = \frac{1232}{308} =4[/tex]

Thus, the kinetic energy increased by a factor of 4.

static friction is acting on stationery object (rest) but kinetic friction acting on a moving body.

The other one is static friction oppose the object to start a motion so it force is great than kinetic friction

Explanation:

Given that,

The force exerted by the stick on the puck is 975 N

The stick is in contact with the puck for 0.0049 s

Initial speed of the puck, u = 0 (at rest)

(a) We need to find the impulse imparted by the stick to the puck.

Impulse = Force × time

J = 4.7775 kg-m/s

(b) Mass of the puck, m = 1.76 kg

We need to find the speed of the puck just after it leaves the hockey stick.

Let the speed be v.

As impulse is equal to the change in momentum.

[tex]J=m(v-u)\\\\4.7775=1.67(v-0)\\\\v=\dfrac{4.7775}{1.67}\\\\v=2.86\ m/s[/tex]

So, when the puck leaves the hockey stick its speed is 2.86 m/s.

Answer :

[tex]\omega=r\ rad/s[/tex] and [tex]f=\dfrac{2}{\pi}\ Hz[/tex]

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s[/tex]

(b) The frequency of oscillation is given by :

[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz[/tex]

Hence, this is the required solution.

Answer:

[tex]400000\ \text{N/C}[/tex]

Explanation:

[tex]q_1[/tex] = Charge at 3000 m = 40 C

[tex]q_2[/tex] = Charge at 1000 m = -40 C

[tex]r_1[/tex] = 3000 m

[tex]r_2[/tex] = 1000 m

k = Coulomb constant = [tex]9\times10^9\ \text{Nm}^2/\text{C}^2[/tex]

Electric field due to the charge at 3000 m

[tex]E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}[/tex]

Electric field due to the charge at 1000 m

[tex]E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}[/tex]

Electric field at the aircraft is [tex]E_1+E_2=40000+360000=400000\ \text{N/C}[/tex].

Answer:

a) For an object with mass M, in a region with a gravitational acceleration g, is:

W = M*g

Then the weight is g times the mass of the object, this means that the weight is not always equal to the mass of the object, this first statement is false.

For example, in Earth the gravitational acceleration is 9.8m/s^2

Then there is no object in Earth with a weight equal to its mass.

b) The force of tension can be the force in a piece of string that is holding an object with mass M.

The force of tension will be always pointing in the direction to the "center" of the string, then it does not push, the tension force "pulls"

The statement is false.

c) we know that the weight is:

W = M*g

This is a force, that for an object that is in the air, will pull the object back to the ground.

Suppose that we also have that object attached to a string, and the object is in the air, now the object starts to fall due to its weight and we also pull down with the string, then the total force pulling down will be the weight plus the tension of the string, then we will have a force larger (in magnitude) than the weight, which means that the statement is false.

Cancer cells are cells that divide relentlessly, forming solid tumors or flooding the blood with abnormal cells. Cell division is a normal process used by the body for growth and repair.

Answer:

105400kgm/s

Explanation:

Momentum is the product of the mass and velocity of an object.

Momentum = mass × velocity

Given

Masses m1 = 1,300 kg and m2 = 1,800

velocity = 34m/s

Magnitude of the total momentum = (m1+m2)v

Magnitude of the total momentum = (1300+1800)((34)

Magnitude of the total momentum = 3100(34)

Magnitude of the total momentum = 105400kgm/s

Answer:

the circulatory system consists of heart, blood, blood vessels.

hope this answer will help you

Answer:

mixture

Explanation:

because of many Ingredients it is called a mixture

Answer:

0.62

Explanation:

we know that [tex]g=\frac{G M}{R^{2}}[/tex]

[tex]\frac{g_{g}}{g_{J}}=\frac{M_{g}}{M_{J}} \times \frac{R_{j}^{2}}{R_{z}^{2}}=\frac{M}{318 M} \times \frac{(11.2)^{2} R_{g}^{2}}{R_{g}^{2}}[/tex]

[tex]\frac{g_{g}}{g_{j}}=\frac{125.44}{318}=0.394[/tex]

We know that  [tex]=\sqrt{\frac{2 h}{g}}[/tex]

here given that each object falls the same distance

[tex]\therefore t \alpha \sqrt{\frac{1}{g}}[/tex]

[tex]\therefore \frac{t_{J}}{t_{B}}=\sqrt{\frac{g_{g}}{g_{J}}}=\sqrt{\frac{g}{0.394}}[/tex]

[tex]\therefore \frac{t_{j}}{t_{g}}=\sqrt{0.394}=0.62[/tex]

Answer:

Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :

Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.

Cat skeleton:  cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.

Answer:

1. 3 points

2. 6 points

3. goalkeeper

4. four quarters

5. eleven players

6. James Naismith

7. Duke nukem ( I'm not sure)

8. no

9. 1936

10. four

HOPE IT'S HELP :)

Answer:

Where the electric potential is constant, the strength of the electric field is zero.

Explanation:

As a test charge moves in a given direction, the rate of change of the electric potential of the charge gives the potential gradient whose negative value is the same as the value of the electric field. In other words, the negative of the slope or gradient of electric potential (V) in a direction, say x, gives the electric field (Eₓ) in that direction. i.e

Eₓ = - dV / dx        ----------(i)

From equation (i) above, if electric potential (V) is constant, then the differential (which is the electric field) gives zero.

Therefore, a constant electric potential means that electric field is zero.

Answer:

The efficiency of the engine is 75.94 %

Explanation:

Given;

mass of the shuttlecraft, m = 2,200 Kg

time required by the shuttlecraft, t = 3 minutes = 180 s

height risen by the shuttle, h = 19 km = 19,000 m

Input power of the shuttle, Pi = 500 kW = 500,000 W

The potential energy due to height risen by the Shuttle on Moon surface is given as;

E = mgh

where;

g is the acceleration due to gravity on moon, = ¹/₆ x 9.81 m/s² = 1.635 m/s²

E = 2200 x 1.635 x 19,000

E = 68343000 J

Output power of the shuttle, is given as  Energy / time

Output power = (68343000) / (180)

Output power = 379,683.33 W

The efficiency of the shuttle is given as;

Efficiency = (Output power) / (Input power)

Efficiency = (379,683.33) / (500,000)

Efficiency = 0.7594

Efficiency(%) = 75.94 %

Therefore, the efficiency of the engine is 75.94 %

Answer:

1.52

Explanation:

Answer: 59.5 deg

Explanation:

I got it right on acellus ✅

Answer:

D. Dwarf Planets

Explanation:

Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.

It is a system that collection of all the planets and spatial bodies revolving around the sun because of the gravitational pull of the sun.

Our Solar System is based on a heliocentric model in which the Sun is assumed to reside at the central point of the planetary system.

As given in the problem Jasmine is late to science class and misses the very beginning of notes for the day. These are Jasmine’s notes: –Round objects that orbit the Sun –Not large enough to have enough gravitational pull to clear the path of their orbit –4 known in the Kuiper Belt (Eris, Haumea, Makemake, and Pluto) –1 in the asteroid belt (Ceres) Jasmine later copied the title of the notes from a friend.

Thus, Jasmine is most likely to copy the Dwarf Planets, therefore the correct answer is option D.

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Answer:

I belive it is the thickness of the wire

Explanation:

There are no other answers and I searched it up and it seems that it is the thickness of the wire.

A current is induced in a wire by moving the wire through a magnetic field. A factor that affects the direction of the current is  the direction in which the wire moves.

Magnetic field is the region of space where an object experiences the magnetic force towards a magnetic material.

The direction of flow of electric current in a current carrying conductor wire depends on many factors. The direction of the wire moves in or out and current will flow towards the direction of a given wire.

Hence, a factor that affects the direction of the current is in the direction in which the wire moves.

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Answer:

There are three main sources of heat in the deep earth:

(1) heat from when the planet formed and accreted, which has not yet been lost;

(2) frictional heating, caused by denser core material sinking to the center of the planet; and (3) heat from the decay of radioactive elements.

Explanation:

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Answer:

h = 61.5m

Explanation:

Initial speed = 34.7m/s

Net force =

F= -mg Sin 10

We are finding distance moved by this skier upon the hill.

ma = - mgSin10

a = -gSin10

g = 9.8

Sin 10 = 0.1564

a = -9.8(0.1564)

a = -1.53

This is the acceleration

(-34.7)² = 2as

= -34.7² = (-2x1.53)s

-1204.09 = -3.06s

We find the value of s

S = -1204.09/-3.06

S = 393.5m

Sin10 = h/393.5m

0.1564 = h/393.5m

h = 0.1564x393.5m

h = 61.5

Answer:

1/3

Explanation:

Gay Lusaac's law states that "the pressure of a given mass of gas is directly proportional with the absolute temperature of the gas, provided that the volume is kept constant."

In formula, we say that

P/T = k

Where

P = pressure at different points

T = temperature at different points

k = constant of proportionality

From the stated formula, if we multiply the temperature by 3, we have

P/3T = k

P * 1/3T = k

And from this, we see the pressure will change by a value of 1/3

Sphere A travels a horizontal distance of (10 m/s) t after time t, while sphere B travels a distance of (5 m/s) t. So sphere B lands on the floor some point between the table's edge and the point X (A).