With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?
a) What will be its speed when it hits the ground?
b) How high does the ball rise?

Answer:

+ 900 J

Explanation:

Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,

ΔE = ΔU

ΔE = Q - W where Q = heat absorbed by system and W = work done by system

Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J

So,  the values of the variables into the equation, we have

ΔE = Q - W

ΔE = + 500 J - (-400 J)

ΔE = + 500 J + 400 J  

ΔE = + 900 J

So, the internal energy change, ΔE = + 900 J

Answer:

A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.

The force will remain the same and is equal to zero.

We have a point charge and a neutral atom.

We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.

The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -

[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]

According to question, we have -

A point charge and a neutral atom.

If initially the distance between the point charge and neutral atom is r meters, then -

q(1) = Q (say)

q(2) = 0   ( Neutral atom has zero charge)

Using Coulomb's law -

[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]

F = 0 Newtons.

Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.

Hence, the force will remain the same and is equal to zero.

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Explanation:

1. Distance, d = 5 m

Time, t = 10 s

Speed = distance/time

[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]

2. Mass, m = 10 kg

Acceleration, a = 3 m/s³

Force, F = mass (m) × acceleration (a)

F = 10 × 3

= 20 N

3. Mass, m = 7 kg

Height, h = 4 m

Potential energy, E = mgh

E = 7 × 9.8 × 4

E = 274.4 J

Hence, this is the required solution.

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

Answer:

[tex]d=79.9m[/tex]

Explanation:

From the question we are told that:

coefficient of static friction [tex]\mu=0.38[/tex]

Velocity [tex]v=87.9=>24.41667m/s[/tex]

Generally the equation for Conservation of energy is mathematically given by

 [tex]\mu*mgd = 0.5 m v^2[/tex]

 [tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]

 [tex]d=79.9m[/tex]

Answer:

wind

Explanation:

just a possible answer.

Answer:

focal length=12cm

Explanation:

object size is equal to 1.00cm

object distance = 18cm

heigh of image = 2.00cm

image distance = ??

but magnification is given by;

M = 2.00/1.00 = 2

but u/v = M

u/18 = 2

u = 36

1/f = 1/u+1/v

1/f = 1/18+ 1/36

1/f = 1/12

f = 12cm

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

so

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

It is Potential energy's

Answer:

Mm, thats the answer trust me men

Answer:

vs = 55.84 m/s

Explanation:

In this case, the source (skydiver with buzzer) is moving towards the observer (friend). The formula for this case will be:

[tex]f' = \frac{v}{v-v_s} f[/tex]

where,

f' = shifted frequency = 2150 Hz

f = actual frequency = 1800 Hz

v = speed of sound = 343 m/s

vs = speed of skydiver = ?

Therefore,

[tex]2150\ Hz = \frac{343\ m/s}{343\ m/s - v_s}(1800\ Hz)\\\\343\ m/s - v_s = \frac{343\ m/s}{2150\ Hz} (1800\ Hz)\\\\v_s = 343\ m/s - 287.16\ m/s\\[/tex]

vs = 55.84 m/s

Answer: The statement [tex]T_{1} > T_{2}[/tex] must be true.

Explanation:

As it is given that heat is being transferred from a substance with temperature [tex]T_{1}[/tex] to a substance with temperature [tex]T_{2}[/tex].

It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.

Hence, in the given situation [tex]T_{1} > T_{2}[/tex].

Thus, we can conclude that the statement [tex]T_{1} > T_{2}[/tex] must be true.

Answer:

The acceleration is a negative as the ball is now moving in the opposite direction.

The velocity would decrease as the ball moves upward

Acceleration remains constant and velocity is negative and decreasing as per the given scenario. The correct option is B.

In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time.

Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.

Velocity is the directional speed of a moving object as an indication of its rate of change in position as observed from a specific frame of reference and measured by a specific time standard.

The rate of displacement of an entity known as its velocity. It is measured in meters per second. The rate of change in velocity of an object is defined as acceleration.

According to the scenario, acceleration remains constant while velocity decreases and is negative.

Thus, the correct option is B.

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Your question seems incomplete, the missing options are:

Answer:

4.60 × 10⁻⁸

Explanation:

From the given information;

Assuming that q charges are transferred, then:

[tex]F = \dfrac{kq^2}{d^2}[/tex]

where;

k = 9 ×10⁹

[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]

[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]

q = 0.012 C

No of the electrons transferred is:

[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]

[tex]= 7.5 \times 10^{16} \ C[/tex]

Initial number of electrons =  N × 47 × no  of moles

here;

[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]

no of moles = 0.0575 mol

∴

Initial number of electrons =  [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]

= 1.63 × 10²⁴

The fraction of electrons transferred  [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]

= 4.60 × 10⁻⁸

Answer:

A.

sửa cho tôi nếu tôi sai

Answer:

ΔT = 25°C

Explanation:

Given that.

The mass of a bullet, m₁ = 30 g = 0.03 kg

The speed of the bullet, v = 900 m/s

Mass of soft iron, m₂ = 1 k

The specific heat of iron, c=490J/kg°C

We need to find the increase in temperature of iron. using the conservation of energy,

Kinetic energy = heat absorbed

[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]

So, the correct option is (A).

Answer:

b is correct.

Explanation:

because of the question you have given

Answer:

(a) 81.54 N

(b) 570.75 J

(c) - 570.75 J

(d) 0 J, 0 J

(e) 0 J  

Explanation:

mass of crate, m = 32 kg

distance, s = 7 m

coefficient of friction = 0.26

(a) As it is moving with constant velocity so the force applied is equal to the friction force.

F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N

(b) The work done on the crate

W = F x s = 81.54 x 7 = 570.75 J

(c) Work done by the friction

W' = - W = - 570.75 J

(d) Work done by the normal force

W'' = m g cos 90 = 0 J

Work done by the gravity

Wg = m g cos 90 = 0 J

(e) The total work done is

Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J  

Answer:

The electric field is given by 4.5 N/C.

Explanation:

Charge density = 80 nC/m3

inner radius, r' = 1 mm

outer radius, r'' = 3 mm

distance,  r = 4 mm

The linear charge density is given by

[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]

The electric field is given by

[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]

Answer:

because density of iron is more than that of water but less then that of Mercury

hope it's helpful

Answer:

What is the potential energy? PE= mghPE= hwKE= 1/2mv2

Answer:1960J

Explanation:

Answer:

Option B

Explanation:

A gaseous, highly reactive non-metal

Answer:B

Explanation:I just took the test

Answer:

Option B. 3.25×10¯¹⁰ mm.

Explanation:

Measurement (cm) = 0.00325×10⁻⁸ cm

Measurement (mm) =?

The measurement in mm can be obtained as follow:

1 cm = 10 mm

Therefore,

0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm

0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm

Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.

The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm

The number given is in standard form and can be written as 3.25*10^-11 cm.

To convert this from centimeter to millimeter, we have to multiply this value by 10.

So, let's 3.25*10^-11 by 10 and get our value in mm

[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]

From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]

Learn more about conversion of units here;

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Answer:

the largest distance we can measure is 10¹⁴ km

Explanation:

Given the data in the question;

Threshold hearing = 10⁻²⁰

smallest distance measured = 1 mm

Largest distance measured will be;

⇒ ( threshold hearing )⁻¹ × smallest distance

= ( 1 / 10⁻²⁰ ) × 1 mm

= 10²⁰ × 1mm

= 10²⁰ mm

we know that; 1000 mm = 10⁶ km

Largest distance = ( 10²⁰ / 10⁶ ) km

= 10¹⁴ km

Therefore, the largest distance we can measure is 10¹⁴ km

Answer:

The speed of projection is 34 m/s.

Explanation:

Height of building, h = 51 m

horizontal distance, d = 74 m

time, t = 8 s

Let the angle is A and the speed is u.

d = u cos A x t

74 = u cos A x 8

u cos A = 9.25 .... (1)

Use second equation of motion

[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]

Squaring and adding both the equations

[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]

Answer:

Explanation:

Here's what we know and in which dimension:

y dimension:

[tex]v_0=30[/tex] m/s

v = 0 (I'll get to that injust a second)

a = -9.8 m/s/s

The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.

x dimension:

Δx = 70 m

v = ??

Velocity is our unknown.

Solving for the time in the y dimension:

[tex]v=v_0+at[/tex] and filling in:

0 = 30 + (-9.8)t and

-30 = -9.8t so

t = 3.1 seconds

We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.

In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.

Δx = vt and

70 = v(6.2) so

v = 11.3 m/s

Explanation:

Given that,

Two resistors of 10 ohms and 15 ohms are connected.

In series combination, the equivalent resistance is given by :

[tex]R_s=R_1+R_2\\\\R_s=10+15\\\\R_s=25\ \Omega[/tex]

In parallel combination, the equivalent resistance is given by :

[tex]\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{15}\\\\R_p=6\ \Omega[/tex]

Hence, this is the required solution.

Answer:

arge number of electrons free to move between the charged ions in the lattice.

Explanation:

The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.

Some important characteristics emerge from this description of the metallic bond:

* It has many free electrons therefore its electrical conductivity is high

* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.

* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity

* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.

Answer:

ΔV = 2 10¹ V

Explanation:

The calculation of the uncertainty or error in an expression is given by

         ΔV = [tex]\frac{dV}{di}[/tex]  |Δi| + [tex]\frac{dV}{dR}[/tex]  |ΔR |

         V = i R

let's make the derivatives

        [tex]\frac{dV}{di}[/tex] = R

        [tex]\frac{dV}{dR}[/tex] = i

we substitute

         ΔV = R | Δi | + i | ΔR |

in the exercise give the values

         i = (5.9 ± 0.4) A

         R = (42.7 ± 0.6) Ω

we calculate

          ΔV = 42.7  0.4 + 5.9  0.6

          ΔV = 20.6 V

          ΔV = 2 10¹ V

the voltage is

         V = i R

         V = 5.9  42.7

          V = 251.9 V

the result is

         V = (25 ± 2) 10¹ V