A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
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Which of these statements best describes the impact of ocean thermal power and current power on the environment?
A. Current power may decrease the fish population.
B. Current power may decrease the gravitational pull of the moon.
C. Ocean thermal power may increase the fish population.
C. Ocean thermal power may increase the gravitational pull of the moon.
Answer:
A. Current power may decrease the fish population.
Explanation:
The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.
The environment is made up of living and non-living components that co-exist and interact with one another.
Harnessing current power from ocean movement will seriously affect the fish population. Most fishes are not sedentary. They move and glide through the water. When current power causes a change in the environment of the fish. This will definitely affect the normal condition prevalent in the body of water.Answer:
A
Explanation:
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A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming free fall conditions, what was the rocket's initial upward velocity?
a. 98.0 m/s
b. 123 m/s
c. 24.5 m/s
d. 49.0 m/s
d. 49.0 m/s
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A football team lost 3 yards on their first play. On the second play, 10 yards were gained. On the third play the quarterback got tackled behind the line of scrimmage and lost 6 yards. What was their net gain or loss after the three plays?
+1yd
Explanation:
Kfgjjh
Answer:
+1 YD
Explanation:
-3 + 10 - 6 = 1
2. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for French class for the third time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The second hallway is filled with students, and she 4covers its 48.0 m length at a speed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0 m length at a speed of 5.00 m/s. How long does it take Suzette to make to class? Did Suzette beat the bell?
Answer:
62 secondsnoExplanation:
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.
A block of mass m1 = 18.5 kg slides along a horizontal surface (with friction, μk = 0.22) a distance d = 2.3 m before striking a second block of mass m2 = 7.25 kg. The first block has an initial velocity of v = 8.25 m/s.
Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
The velocity of block 2 and the distance traveled by it prior to being at rest post-collision are 19.5 m/s and 87.8 m. Check the calculations below:
FrictionGiven that,
[tex]m_{1}[/tex] = 18.5 kg
d = 2.3m
To find,
Acceleration of block 1:
∑[tex]F = ma[/tex]
⇒ -m₁gμ = m₁a
⇒ a = -gμ
⇒ a [tex]= -(9.8 m/s^2) (0.22)[/tex]
∵ a [tex]= -2.16 m/s^2[/tex]
Now,
To determine the velocity of block one prior to the collision:
We know,
The initial velocity of block 1 = 8.25 m/s
⇒ [tex]v^2 = v_{o}^2 + 2[/tex]aΔx
⇒ [tex]v^2 = (8.25 m/s)^2 + 2 (-2.16 m/s^2) (2.3 m)[/tex]
∵ [tex]v = 7.63 m/s[/tex]
We also know,
[tex]m_{2}[/tex] = 7.25 kg
Now,
The velocity of block 2 post collision:
⇒ [tex]m_{1} u_{1} + m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}[/tex]post-collision
Through this,
⇒ [tex](18.5 kg) (7.63 m/s) = (7.25 kg) v[/tex]
∵[tex]v = 19.5 m/s[/tex]
The distance can be found through:
⇒ [tex]v^2 = v_{o} ^{2} + 2[/tex][tex]a[/tex]Δ[tex]x[/tex]
⇒ [tex](0 m/s)^2 = (19.5 m/s)^2 + 2 (-2.16 m/s^2)[/tex]Δ[tex]x[/tex]
∵ Δ[tex]x = 87.8 m[/tex]
Thus, 19.5 m/s and 87.8 m are the correct answers.
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Open Box. Consider a hollow box with the top missing. The sides have negligible thickness and each has length L and mass m. (a) Find the x-coordinate of the center of mass.
Answer:
x_{cm} = L / 2
Explanation:
The center of mass is defined by
[tex]x_{cm}[/tex] = 1 / M ∑ m_{i} x_{i}
where M is the total mass of the system
in this case the system is continuous so, for which we use the density
ρ = dm / dx
dm = ρ dx
substituting
x_{cm} = 1 / M ∫ x ρ dx
x_{cm} = ρ / M ∫ x dx
we integrate and evaluate from x = 0 to x = L
x_{cm} = ρ / M (L² /2 -0)
we introduce the density which is constant
ρ = M / L
x_{cm} = 1 /M (M/L) L² / 2
x_{cm} = L / 2
A man walking at 3.56 m/s accelerates at 2.50 m/s2 for 9.28 s. How far does he get?
141 m
26.8 m
44.6 m
248 m
Question 16 1 pts Jessie feels pressured by his parents to get a job. This is an example of the law of?
readiness
disuse
effect
belonging
An increase in temperature the kinetic energy and average speed of the gas particles. As a result, the pressure on the walls of the container . Answer Bank What temperature must a gas, initially at 10 ∘C, be brought to for the pressure to triple?
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]
(A) Electricity and Magnetism
A). Three point charges are aligned along the x axis as shown in
Fig. Find the electric field at (a) the position (2, 0) and (b) the
position (0, 2).
electricity
Explanation:
the position (2,o
A car takes 8 s to increase its velocity from 10 m/s to
30 m/s. What is its average acceration?
Answer:
Average acceleration is 2.5 m/s^2.
Explanation:
Average acceleration can be found by dividing the change in speed by the change in time.
We know that the initial velocity=10m/s, the final velocity=30m/s and time elapsed is t=8s.
a= (Final velocity-Initial Velocty)/t
a=(30-10)/8=20/8=2.5 m/s^2
Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.
Answer:
F=1.3x10^-2N
Explanation:
Fe= k(6x10^-7C)^2/(0.5)^2
Electrical force of attraction between the balloons is F=1.3x10^-2N
The electric force of attraction between two balloons should be F=1.3x10^-2N.
Calculation of the electric force;Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.
So, here the electric force is
Fe= k(6x10^-7C)^2/(0.5)^2
F=1.3x10^-2N
hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.
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A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and cresting a 2800 m pass. (a) Neglecting friction, what should the speed of the car be at the top of the second pass? (b) Find the actual speed of the car if the work due to nonconservative forces is – 5 x106 J.
Answer:
a) v = 88.54 m/s
b) vf = 26.4 m/s
Explanation:
Given that;
m = 1400.0 kg
a)
by using the energy conservation
loss in potential energy is equal to gain in kinetic energy
mg × ( 3200-2800) = 1/2 ×m×v²
so
1400 × 9.8 × 400 = 0.5 × 1400 × v²
5488000 = 700v²
v² = 5488000 / 700
v² = 7840
v = √7840
v = 88.54 m/s
b)
Work done by all forces is equal to change in KE
W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)
we substitute
1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )
488000 = 700 vf²
vf² = 488000 / 700
vf² = 697.1428
vf = √697.1428
vf = 26.4 m/s
A curve that has a radius of is banked at an angle of . If a -kg car navigates the curve at without skidding, what is the minimum coefficient of static friction between the pavement and the tires
Answer:
0.65
Explanation:
For whatever reasons, the parameters are not giving. So, I will assume by myself to make the calculations easier. You can substitute whatever it is to it from your question.
Given that
Radius of the road, r = 63 m
Speed of the car, v = 20 m/s
The relationship between a car that is passing through a curve and it's frictional force is said to be
U(s) * g = v²/r
In the formula above,
U(s) = coefficient of static friction
g = acceleration due to gravity, 9.8 m/s²
v = velocity of the car
r = radius of the road
Now when we substitute the earlier stated values, we have
U(s) * 9.8 = 20² / 63
U(s) * 9.8 = 400 / 63
U(s) * 9.8 = 6.35
U(s) = 6.35 / 9.8
U(s) = 0.65
Thus, our coefficient of static friction, based on the stated values is 0.65
A spring gun is able to launch a 7.0 gram marble to a vertical height of 22 mmeasured from the compressed point of the marble). If the spring iscompressed 8.0 cm from its relaxed state, what will be the spring constant
Answer: Spring constant = 472N/m
Explanation:
The change in gravitational potential energy by the spring is given as = mgh
where m= 7.0 g = 7 X 10 -3kg
g= 9.8m/s
h= 22m
Gravitational potential energy= mgh
= 7.0 x 10^-3 X 9.8 x 22 = 1.5092 J
Remember that change in gravitational potential energy by the spring =elastic potential energy
Therefore, Potential energy P. E = 1/2 K x²
where K= CONSTANT
x= 8.0
2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m
2. For a rotating rigid body, which of the following statements is NOT correct?
a. All points along a rotating rigid body move with constant linear speed.
b. Points along a rotating body have same angular velocities.
c. Points along a rotating body move through the same angle in equal time intervals.
d. All points have the same angular acceleration.
Answer:
dasgfwe
Explanation:
Isaac walks 4 blocks north. Then he turns around and walks 1 block south.
Which of the following correctly describes Isaac's motion?
A. Isaac walked a distance of 5 blocks, and his displacement was 3
blocks north.
B. Isaac walked a distance of 5 blocks, and his displacement was 5
blocks.
C. Isaac walked a distance of 3 blocks, and his displacement was 3
blocks north.
D. Isaac walked a distance of 3 blocks north, and his displacement
was 5 blocks.
Answer:
Isaac walked a distance of 5 blocks, and his displacement was 3 blocks north.
Explanation:
Distance is what he covered from the beginning, while displacement was what he covered in a specific direction
How much would a 15.0 kg object weigh on that planet? Round the answer to the nearest whole number.
Answer:
168
Explanation:
Answer: a 15 kg object would weigh the most on Neptune
168 N
50 POINTS!!! PLEASE!!!!!! Pls help me with this been stuck on it in like FOREVER!! PLSS.
Answer:
it will move in 356.5g in diagonal direction(resultant or between the both direction)
Explanation:
285+428/2=356.5g
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Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD.
Answer:
KE= 1/2mv²
Explanation:
The kinetic energy of a body is the energy possessed by virtue of the body in motion
Given the parameters
m which is the mass of the body
v which is the velocity of the body too
K.E = kinetic energy
The expression for the kinetic energy of a body is given as
KE= 1/2mv²
A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.
Answer:
xn = 36.28 [m]
Explanation:
To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.
[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]
where:
x - xo = displacement of the car [m]
Vo = initial velocity = 0
t = time = 7.2 [s]
a = acceleration = 1.4 [m/s^2]
The initial velocity is zero, as the car begins its movement from rest.
xn = (x - xo), Now replacing
xn = (0*7.2) + 0.5*1.4*(7.2^2)
xn = 36.28 [m]
Describe four ways in which improving your muscular fitness would have a positive affect in your life
Four ways in which improving our muscular fitness would have a positive affect on our lives are given below:
Injury preventionBody compositionChronic disease prevention.Lifetime muscle and bone healthMuscular fitnessMuscle fitness simply means having muscles that can lift heavier objects or muscles that will work longer before becoming exhausted.
It helps us improves when a person does activities that build or maintain muscles or that increase how long a person can use his or her muscles.
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Magnets are usually made up of which material
A. plastic
B. iron ore
C. copper
D. gold
Answer:
B. iron ore
Explanation:
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Can we cure aging? Support your answer?
Answer: yes it can
Explanation:
Answer:
Yes you can. You need to eat healthy and have the proper diet and always exercise and build muscles, this can slow down aging, and most importantly, we must always be calm. However there is no way to stop aging for that’s part of life.
Explanation:
21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?
Answer:
v = 66 m/s
Explanation:
Given that,
The initial velocity of a car, u = 0
Acceleration of the car, a = 11 m/s²
We need to find the final velocity of the toy after 6 seconds.
Let v is the final velocity. It can be calculated using first equation of motion. It is given by :
v = u +at
v = 0 + 11 m/s² × 6 s
v = 66 m/s
So, the final velocity of the car is 66 m/s.
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression
Answer:
(a) The work done by the gas on the surroundings is, 17537.016 J
(b) The entropy change of the gas is, 73.0709 J/K
(c) The entropy change of the gas is equal to zero.
Explanation:
(a) The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 4 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 240 K
= initial volume of gas =
= final volume of gas =
Now put all the given values in the above formula, we get:
The work done by the gas on the surroundings is, 17537.016 J
(b) Now we have to calculate the entropy change of the gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
Thus, w = q = 17537.016 J
Formula used for entropy change:
The entropy change of the gas is, 73.0709 J/K
(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.
As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0
So, from this we conclude that the entropy change of the gas must also be equal to zero.
Explanation:
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
Please help me! What is Ohm's law?
Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:
voltage = current x resistance
This formula tells you that current and resistance is the voltage of an energy bond.
Hope this helps you!
Answer: What is Ohm’s Law?
Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.
E = I × R
It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.
Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.
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The speed of a car is decreasing from 35 m/s to 15 m/s in 4s