Answer:
F = 6[N].
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
[tex]P=m*v\\or\\P=F*t[/tex]
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 10 [kg]
v = velocity [m/s]
F = force [N]
t = time = 5 [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
[tex](m_{1}*v_{1})+F*t=(m_{1}*v_{2})[/tex]
where:
m₁ = mass of the object = 10 [kg]
v₁ = velocity of the object before the impulse = 1 [m/s]
v₂ = velocity of the object after the impulse = 4 [m/s]
[tex](10*1)+F*5=10*4\\10+5*F=40\\5*F=40-10\\5*F=30\\F=6[N][/tex]
A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
A block of mass, m, sits on the ground. A student pulls up on
the block with a tension, T, but the block remains in contact
with the ground. What is the normal force on the block?
Answer a
Explanation: a
The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
How would the mass and weight of an object on the Moon compare to the mass and weight of the same object on Earth? * Mass and weight would both be less on the Moon. Mass would be the same but its weight would be less on the Moon. Mass would be less on the Moon and its weight would be the same. Mass and weight would both be the same on the Moon.
Answer:
B. Mass would be the same but its weight would be less on the Moon.
Explanation:
The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
Thus, the mass of a body is constant either on the Earth or on the Moon. But the weight would be less on the Moon because the gravitational force on the Moon is far less than that on the Earth. Therefore the weight would be less on the Moon.
The appropriate option is B.
The mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
The prime focus to solve this problem is the mass and weight of an object. The mass of a body can be expressed as the quantity of matter it contains. While the weight of a body is the extent of the gravitational force impressed on the body by a massive body.
So, the mass of a body is constant either on the Earth or on the Moon. But the weight of an object will depend on the mass and the gravitational acceleration.
W = mg
Here, W is weight, m is mass and g is gravitational acceleration.
Weight would be less on the Moon because the gravitational force on the Moon is far less (due to lower value of g) than that on the Earth. Therefore the weight would be less on the Moon.
Thus, we can conclude that the mass will remain same on both moon and Earth, but weight will be lesser on Moon than Earth. Hence, option (B) is correct.
Learn more about the mass and weight here:
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On Venus, the atmospheric temperature is a hot 720 K due to the greenhouse effect. It consists mostly of carbon dioxide (molar mass 44 g/mol) and the pressure is 92 atm. What is the total translational kinetic energy of 3 moles of carbon dioxide molecules?
Answer:
The value is [tex]E_t = 17958.2 \ J[/tex]
Explanation:
From the question we are told that
The atmospheric temperature is [tex]T_a = 720 \ K[/tex]
The molar mass of carbon dioxide is [tex]Z = 44 \ g/mol[/tex]
The pressure is [tex]P = 92 \ atm =[/tex]
The number of moles is [tex]n = 3 \ moles[/tex]
Generally the translational kinetic energy is mathematically represented as
[tex]E_t = \frac{f}{2} * n * R T[/tex]
Here R is the gas constant with value [tex]R = 8.314 J\cdot K^{-1}\cdot mol^{-1}[/tex]
Generally the degree of freedom of carbon dioxide in terms of translational motion is f = 3
So
[tex]E_t = \frac{ 3}{2} * 2 * 8.314 * 720[/tex]
=> [tex]E_t = 17958.2 \ J[/tex]
A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?
Answer:
Angular speed = 27.78 rad/s (Approx)
Explanation:
Given:
Diameter = 21.6 cm
Speed = 3 m/s
Find:
Angular speed
Computation:
Radius = 21.6 / 2 = 10.8 cm = 0.108 m
Angular speed = v / r
Angular speed = 3 / 0.108
Angular speed = 27.78 rad/s (Approx)
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
A rigid tank contains an ideal gas at 300 kPa and 600 K. Now half of the gas is withdrawn from the tank and the gas is found at 100 kPa at the end of the process. Determine (a) the final temperature of the gas and (b) the final pressure if no mass was withdrawn from the tank and the same final temperature was reached at the end of the process.
A stretched string is observed to have four equal segments in a standing wave driven at a frequency of 480 Hz. What driving frequency will set up a standing wave with five equal segments?
a) 360 Hz.b) 240 Hz.c) 600 Hz.d) 120 Hz.
Answer:
C) 600 Hz
Explanation:
The fundamental frequency can be related to the driving frequency by the expression below;
f(n) = n * f(1)
Where f(1)= fundamental frequency
f(n) = driving frequency
There are four equal segments in the standing wave , then our n= 4 and our f(n)=4, then we can get the fundamental frequency here
f(4) = 4× f(1)
480 = 4× f(1)
f(1) = 480/4
f(1)=120Hz
Hence, fundamental frequency is 120Hz
To calculate the driving frequency that will set up a standing wave with five equal segments?
n=5
f(n) = n× 120Hz
f(5) = 5×120Hz
= 600Hz.
Hence, the driving frequency that will set up a standing wave with five equal segments is 600Hz
explain an experiment of the phenomenon of rainfall
Unclear/incomplete question. However, I inferred you need an explanation of the phenomenon of rainfall.
Explanation:
Basically, the phenomenon of rainfall follows a natural cycle called the water cycle. What we call 'rainfall' occurs when water condensed (in liquid form) in the atmosphere is made to fall down on the ground as tiny droplets as a result of the forces of gravity.
The water cycle makes rainfall possible:
First, water on the earth's surface is evaporated (or is absorbed into) the atmosphere.Next, it then condensed into liquid form; which later falls to the surface to the ground again. And the process continues.A radioactive nuclide of atomic number Z emits an alpha particle and the daughter nucleus then emits a beta particle. What is the atomic number of resulting nuclide?
A) Z-1
B) Z+1
C) Z-2
D) Z-3
Answer:
A) Z-1
Explanation:
when a radioactive element of atomic number Z emits an alpha particle, the mass of the new nucleus decreases by 2, i.e the new atomic number of the element = ( Z- 2).
Also, when the daughter nucleus emits a beta particle, the new nucleus increases by 1, that is the new atomic number of the element = (Z + 1).
Thus, the atomic number of resulting nuclide = Z ( - 2) + ( + 1).
= Z - 2 + 1
= Z - 1
Therefore, the atomic number of resulting nuclide is Z - 1
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
Learn more about acceleration here: https://brainly.com/question/605631
A heat pump has a coefficient of performance of 3.85 and operates with a power consumption of 7020 W. How much energy does it deliver into a home during 1 h of continuous operation?
Answer:
97.3 MJ
Explanation:
The formula for the coefficient of Perfomance is given as
COE = Q/W, where
COE is the coefficient of Perfomance
Q is the heat provided
W serves as the work input.
Dividing both sides of the equation by a factor of time t, we get the coefficient of Perfomance in terms of heating power and input power, so we say
COE = P / P(i),
making heating power, P the subject of formula, we have
P = COE * P(i)
P = 3.85 * 7020 * 1 * 3600
P = 97297200 J
P = 97.3 MJ