Answer: SORRY
SORRY AM DOING IT FOR POINTS
Explanation:
a car is traveling with a velocity of 40 m/s and has a mass of 1120kg the car has kinetic energy
(TCO 6) A car travelling at 70 kilometers per hour hits a block wall and comes to a complete stop. If the time for the car to reach a complete stop is 450 ms and the wall does not move, how much force was exerted on the car? The mass of the car is 1500 kg.
Answer:
F = 64800 N
Explanation:
Given that,
Initial speed of a car, u = 70 km/h = 19.44 m/s
Finally it comes to a stop, v = 0
Time,t = 450 ms
The mass of the car is 1500 kg
We need to find the force exerted on the car. The force exerted on an object is given by :
F = ma
So,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1500\times (19.44-0)}{450\times 10^{-3}}\\\\F=64800\ N[/tex]
So, the required force is equal to 64800 N.
A car travelling at 25 m/s has momentum 20,000 Kgm/s, calculate the mass of the car.
[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]
We have ―
Velocity of the car, v = 25 m/sMomentum of the car, P = 20,000 kg.m/sWe have been asked to calculate the mass of the car, m.
[tex]\qquad\implies\boxed{\red{\sf{ P = mv}}}\\[/tex]
P denotes momentumm denotes massv denotes velocity[tex] \quad \twoheadrightarrow\sf { 20000 = 25m} \\ [/tex]
[tex] \quad \twoheadrightarrow\sf { \cancel{\dfrac{20000}{25}} = m} \\ [/tex]
[tex]\quad\twoheadrightarrow\boxed{\red{\sf{ m = 800 \; kg}}}\\[/tex]
Therefore, mass of the car is 800 kg.
An object has 4J of kinetic energy and 16J of potential energy. How much mechanical energy does it have?
A. 64J
B. 12J
C. 20J
D. 4J
Answer: C. 20J
Explanation: im pretty sure sorry if im wrong :(
describe the human condition before Science and technology was practice.
A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This force causes the house to move from rest to an upward speed of 15 cm/s in 5.0 s. What is the mass of the house?
Answer:
m = 95000 kg
Explanation:
Given that,
Net force acting on the house, F = 2850 N
Initial speed, u = 0
Final speed, v = 15 cm/s = 0.15 m/s
We need to find the mass of the house. Let the mass be m. We know that the net force is given by :
F = ma
Where
a is the acceleration of the house.
So,
[tex]F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg[/tex]
So, the mass of the house is equal to 95000 kg.
The speed of a wave is 1500 m/s. What is the frequency of the wave if the wavelength is 2 m?
O 1502 Hz
O 750 Hz
O 375 Hz
O 300 Hz
Answer:
750 Hz
Explanation:
speed = freq * wl
1500 = f * 2
f = 750 Hz
A.) A gemstone of mass 1.8 kg compresses a scale's spring by 2.6 cm.
Determine the spring constant.
B.) How much would the spring in the previous question compress if a 5.2 kg mass was placed on the scale?
Explanation:
Given that,
Mass, m = 1.8 kg
Compression, x = 2.6 cm
We know that,
Force on spring = weight
So,
[tex]mg=kx[/tex]
Where
k is spring constant
[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{1.8\times 9.8}{2.6\times 10^{-2}}\\\\k=678.46\ N/m[/tex]
(2) If m = 5.2 kg
[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{1.8\times 9.8}{678.46}\\\\x=2.6 \ cm[/tex]
Hence, this is the required solution.
A pendulum with a mass of 250 g is released from its maximum displacement of 5 degrees and swings with a simple harmonic motion. If the period is 4 seconds, calculate a) the length of the pendulum, b) the amplitude, c) frequency, d) maximum velocity, e) total energy, f) maximum height.
(a) The length of the pendulum is 3.96 m.
(b) The amplitude is 7.89 m.
(c) The frequency is 0.25 Hz.
(d) The maximum velocity is 12.38 m/s.
(e) The total energy is 9.67 J.
(f) The maximum height is 3.94 m.
Length of the pendulum
The length of the pendulum at the given parameters is determined using the following formula.
[tex]T = 2\pi \sqrt{\frac{l}{gcos(\theta)} } \\\\\frac{T}{2\pi } = \sqrt{\frac{l}{gcos(\theta)} }\\\\\frac{T^2}{4\pi ^2 } =\frac{l}{gcos(\theta)}\\\\l = \frac{gcos(\theta)T^2}{4\pi^2} \\\\l = \frac{(9.8)cos(5)\times (4)^2}{4\pi^2} \\\\l = 3.96 \ m[/tex]
Amplitude of the pendulumThe amplitude of the pendulum is calculated as follows;
mgLcosθ = ¹/₂FA
mgLcosθ = ¹/₂(mg)A
Lcosθ = ¹/₂A
A = 2Lcosθ
A = 2 x 3.96 x cos(5)
A = 7.89 m
Angular speed of the wave[tex]\omega = \sqrt{\frac{g}{l} } \\\\\omega = \sqrt{\frac{9.8}{3.96} } \\\\\omega = 1.57 \ rad/s[/tex]
Frequency of the waveω = 2πf
f = ω/2π
f = (1.57) / (2π)
f = 0.25 Hz
Maximum velocityThe maximum velocity is calculated as follows;
v = Aω
v = 7.89 x 1.57
v = 12.38 m/s
Total energyE = mgLcosθ
E = 0.25 x 9.8 x 3.96 x cos(5)
E = 9.67 J
Maximum height reached by the pendulumh = Lcosθ
h = 3.96 x cos(5)
h = 3.94 m
Learn more about length of pendulum here: https://brainly.com/question/8168512
You leave a pastry in the refrigerator on a plate and ask your roommate to take it out before youget home so you can eat it at room temperature, the way you like it. Instead, your roommateplays video games for hours. When you return, you notice that the pastry is still cold, but thegame console has become hot. Annoyed, and knowing that the pastry will not be good if it ismicrowaved, you warm up the pastry by unplugging the console and putting it in a clean trashbag (which acts as a perfect calorimeter) with the pastry on the plate. After a while, you find thatthe equilibrium temperature is a nice, warmTeq.. You know that the game console has a mass ofm1. Approximate it as having a uniform initial temperature ofT1. The pastry has a mass ofm2and a specific heat ofc2, and is at a uniform initial temperature ofT2. The plate is at the sameinitial temperature and has a mass ofm3and a specific heat ofc3. What is the specific heat ofthe console
Answer:
[tex]c_{e1} = \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \ (T_{Teq} - T_2) }{m_1 (T_1 - T_{eq}) }[/tex]
Explanation:
This is a calorimeter problem where the heat released by the console is equal to the heat absorbed by the cupcake and the plate.
Q_c = Q_{abs}
where the heat is given by the expression
Q = m c_e ΔT
m₁ c_{e1) (T₁-T_{eq}) = m₂ c_{e2} (T_{eq} -T₂) + m₃ c_{e3} (T_{eq}- T₁)
note that the temperature variations have been placed so that they have been positive
They ask us for the specific heat of the console
[tex]c_{e1} = \frac{(m_2 c_{e2} \ + m_3 c_{e3} ) \ (T_{Teq} - T_2) }{m_1 (T_1 - T_{eq}) }[/tex]