What is the volume of an object if it has a mass of 10 grams and a density of 87 g/ml

Answer:

300000 Joules

Explanation:

Recall that the unit Watts is Joules per second, derived from the quotient of energy per unite of time. Therefore, to calculate the energy used on the given time, we need to multiply the power used (1000 W) times the time used. Then we need to express the 5 minutes in seconds to get our answer in Joules:

5 minutes = 5 * 60 seconds = 300 seconds

Final energy calculation gives:

E = 1000 W * 300 sec= 300000 Joules

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Answer:68.15m/s

Explanation:

Given:

v₁=15m/s

a=6.5m/s²

v₁=?

x=340m

Formula:

v₁²=v₁²+2a (x)

Set up:

=[tex]\sqrt{15m/s} ^{2} +2(6.5m/s^2)(340m)[/tex]

9514 1404 393

Answer:

  Not Necessarily

Explanation:

If the object is changing speed or direction, then it is accelerating. If it is maintaining the same speed and direction, it is not accelerating.

Answer:

the acceleration of gravity.

Answer:

g stand for the acceleration of gravity .

Explanation:

Answer:

His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.

Explanation:

According to Einstein's 1915 general theory of relativity, the force of gravity arises from the curvature of space and time.

According to theory of special relativity:

1. The laws of physics are the same for all non-accelerating observers

2. The speed of light in a vacuum was independent of the motion of all observers.

His general theory includes uniform and accelerated motion, but his special theory applies only to uniform motion.

Answer:

for those who dont like to read

the answer is A.

hope i helped

Explanation:

Answer:

2.83 m/s

Explanation:

you have the right answer

Answer:

options C is correct

Explanation:

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Answer:

option c should be the answer

Answer:

[tex]a=2.4\ m/s^2[/tex]

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

[tex]d=ut+\dfrac{1}{2}at^2[/tex]

or

[tex]d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2[/tex]

So, the acceleration of the car is [tex]2.4\ m/s^2[/tex].

Answer:

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

Explanation:

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Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 40 × 4

We have the final answer as

Hope this helps you

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

The stopping distance of car achieved during the braking is of 45.56 m.

Given data:

The initial speed of car is, u = 27 m/s.

The final speed of car is, v = 0 m/s. (Because car comes to stop finally)

The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].

In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.

Therefore,

[tex]v^{2}=u^{2}+2(-a)s[/tex]

Here, s is the stopping distance.

Solving as,

[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]

Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.

Learn more about the kinematic equation of motion here:

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Explanation:

The velocity of car A relative to car B is (10km/h+40km/h)=50km/h

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m  * 1609.344 m

  = 149.668 * 10^8 m

next we will express the distance between the earth and the sun

[tex]r = \frac{a(1-E^2)}{1+Ecos\beta }[/tex]   --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

[tex]\beta[/tex] = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

[tex]v^2 = \frac{4\pi^2 }{r_{c} }[/tex]   ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

([tex]V^2 = (\frac{4\pi^{2} }{149.626*10^9})[/tex]

therefore : V = 1.624* 10^-5 m/s

Answer:

9.57m/s²

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity  = 90m/s

Distance  = 423m

Unknown:

Acceleration of the race car  = ?

Solution:

To solve this problem, we should apply one of the appropriate motion equations;

      V²  = U²   + 2as

Where V is the final velocity

           U is the initial velocity

           a is the acceleration

           s is the distance

  90²  = 0²  + 2 x a x 423

  8100 = 846a

      a  = 9.57m/s²

Answer:

i needed points it was an emergency sorry

Explanation:

Answer:

Kindly check the explanation section.

Explanation:

For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.

From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.

Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.

Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).

If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.

The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500

The fracture strength = .75 × Ah × Fhb = 309 kips.

The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.

Answer:

Covalent Bond

Explanation:

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Answer: Electrical

Explanation: Atoms are tied together by electrical bonding forces.

Answer:

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

Explanation:

For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.

Let's start with the projectile launch

as the body leaves the vertical its velocity must be horizontal

         x = v₀ₓ t

         y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero

         t = √ 2 y₀ / g

we substitute

        x = vox √2y₀ / g

        v₀ₓ = √(g / 2y₀)     x

In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)

       v₀ₓ = √(g /(2 (H -L))    D

this is the minimum speed to cross the well.

Now let's use conservation of energy

starting point. On the platform

      [tex]Em_{o}[/tex] = U = m g H

final point. At the bottom of the swing

      Em_{f} = K + U = 1 / 2m v² + m g (H -L)

as there is no friction the mechanical energy is conserved

        Em_{o} = Em_{f}

       m g H = 1 / 2m v² + m g (H -L)

        v = √ (2gL)

let's write our two equations

the minimum speed to cross the well

       v₀ₓ = √ (g /(2 (H -L))    D

the speed at the bottom of the oscillatory motion

       v = √ (2g L)

we analyze the extreme cases

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well

      V₀ₓ = v

      g / (2 (H -L) D² = 2g L

       4 L (H- L) = D²

        4 H L - 4 L2 - D² = 0

        L² - H L - D² / 4 = 0

let's solve the quadratic equation

      L = [H ± √ (H2-D2)] / 2

we assume that H> D

       L = ½ H [1 + - RA (1 - (D / H) 2)]

The two values ​​of La give the range of values ​​for which the two speeds are equal

A) The person lands in the moat if the rope's length is very short because :

B) The person lands in the moat if the rope length is similar to the height of the platform because :

Following the assumptions;

size of the person is much smaller than L and H

D = horizontal distance

Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed  and  The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.

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Answer:

Capillary Action

Explanation:

Answer:

They turn off

Explanation:

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The expression is  [tex]W_c =  P_o V_o ln (R_v)[/tex]  

Explanation:

Generally smallest workdone done by  a gas is mathematically represented as

          [tex]dW  =  PdV[/tex]

Generally for an isothermal process

    [tex]PV  =  nRT = constant [/tex]

=>   [tex]P = \frac{nRT}{V}[/tex]

Generally the total workdone is mathematically represented as

   [tex]W_c =  \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]

=> [tex]W_c = nRT  \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]

=>  [tex]nRT [lnV]   | \left \ {V_f}} \atop {V_o}} \right.[/tex]

=>  [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]

=>  [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]

From the question [tex]\frac{V_f}{V_o }  =  R_v[/tex]

=> [tex]W_c =  P Vln (R_v)[/tex]

at initial  state

[tex]W_c =  P_o V_o ln (R_v)[/tex]  

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Answer:

Explanation:

Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is as seen below

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ coloumbs.

The required value of effective charge on HF molecule, due to H and F is  1.602 × 10⁻¹⁹ Coulombs.

The given problem is based on the concept of effective charges. The net positive charge carried out by the electrons of atomic species, after forming a polyelectronic atom  is known as Effective charge.

As per the given problem, the Hydrogen fluoride (HF) is an ionic/electrovalent compound that dissociates into ions when dissolved in water. It's dissociation is given as,

HF ⇄ H⁺ + F⁻

There is a transfer of electron from the hydrogen atom which produces the hydrogen ion (H⁺), while the fluorine atom receives the donated ion to become negatively charged (F⁻). The amount of charge in one electron is generally given as 1.602 × 10⁻¹⁹ Coulombs.

Thus, we can conclude that the required value of effective charge on HF molecule, due to H and F is 1.602 × 10⁻¹⁹ Coulombs.

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Answer:

did you get the answer???

Answer: 38 m/s north

Explanation:

Answer:

Explanation:

The total travel time Suzette experiences is the sum of the times in each hallway. Using

  time = distance/speed

we can add the times.

  (35.0 m)/(3.50 m/s) +(48.0 m)/(1.20 m/s) +(60 m)/(5.0 m/s)

  = 10 s + 40 s + 12 s

  = 62 s

It takes Suzette 62 seconds to get to class. She does not beat the bell.

By cooling down the food, the thermal energy in the food molecules is reduced.

A refrigerator is an appliance that is commonly used in the home for the purpose of cooling down a substnace especially water and drinks.

Due to the fact that the molecules that compose matter are is in a state of constant random motion, the food molecules contain thermal energy. Hence, by cooling down the food, the thermal energy in the food molecules is reduced.

Learn more about thermal energy: https://brainly.com/question/11278589

Answer:

a

The pressure will increase

b

[tex]T_2 =  576^oC[/tex]

Explanation:

From the ideal gas law we have that

     [tex]PV  =  nRT[/tex]

We see that the temperature varies directly with the pressure so if there is an increase in temperature that pressure will increase

   The initial  temperature is [tex]T_i  =  10^oC = 10 + 273 =  283 \  K [/tex]

The objective of this solution is to obtain the temperature of the gas where the pressure is tripled

Now from the above equation given that nR and V  are constant  we have that

    [tex]\frac{P}{T}  =  constant[/tex]

=>  [tex]\frac{P_1}{T_1}  =\frac{P_2}{T_2}[/tex]

Let assume the initial  pressure is [tex]P_1 =  1 Pa[/tex]

So tripling it will result  to the pressure being [tex]P_2 =  3 Pa[/tex]

So

     [tex]\frac{1}{283}  =\frac{3}{T_2}[/tex]  

=>   [tex]T_2  =  3 *  283[/tex]

=>    [tex]T_2  =  3 *  283[/tex]

=>    [tex]T_2  = 849 \ K [/tex]

Converting back to [tex]^oC[/tex]

   [tex]T_2  =  849 -  273[/tex]

=>  [tex]T_2 =  576^oC[/tex]

Answer:

1, 5, 4, 6

Explanation:

bc