Answer:
Một ô tô có khối lượng m=1000kg đang chạy với vận tốc 18km/h thì hãm phanh.Biết lực hãm là 2000N. Tính quãng đường xe còn chạy thêm trước khi dừng ... Chiếu phương trình của định luật II Newtơn mà →F=m. ... chuyển động ta có F=ma, suy ra gia tốc chuyển động của xe ( với F=2000N) ...
Explanation:
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?
a. All of the mechanical energy is converted into other forms
b. Some of the mechanical energy is converted into other forms
c. No mechanical energy is converted into other forms
In a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
In a perfect elastic collision, both momentum and kinetic energy of the particles are conserved.
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
[tex]\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2 ^2= \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2[/tex]
When the collision is not perfectly elastic, only momentum is conserved but the kinetic energy is not conserved.
Thus, we can conclude that in a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
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You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?
Answer:
v₀ = 2.67 m / s
Explanation:
This problem can be solved using the Kinetic Enemy Work Theorem
W = ΔK
Work is defined by the relation
W = fr. d
The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.
the friction force opposes the displacement therefore its angle is 180º
W = - fr d
Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane
Y axis
N- Wy = 0
N - W cos tea = 0
the friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
W = - μ W cos θ d
let's look for kinetic energy
the minimum velocity at the highest point is zero
K_f = 0
the initial kinetic energy is
K₀ = ½ m v₀²
we substitute energy in the work relationship
- μ W cos θ d = 0 - ½ m v₀²
v₀² = - μ W cos θ 2d / m
Let's use trigonometry to find distance d
sin θ= y / d
d = y /sin θ
d = 3.50 / sin 30
d = 7 m
let's calculate
v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50
v₀ = √7.129
v₀ = 2.67 m / s
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
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If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
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
A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run though the wire? A from the time that the magnet is pushed into the coil to the time it is pulled out B while the magnet remains within the coil C while the magnet is moving D only while the magnet is being pulled out of the coil
Answer:
C. while the magnet is moving
Explanation:
Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.
In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.
The current is induced because of the motion involved. Thus, the appropriate option is C.
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
A 92-kg man climbs into a car with worn out shock absorbers, and this causes the car to drop down 4.5 cm. As he drives along he hits a bump, which starts the car oscillating at an angular frequency of 4.52 rad/s. What is the mass of the car ?A) 890 kg
B) 1900 kg
C) 920 kg
D) 990 kg
E) 760 kg
Answer:
the mass of the car is 890 kg
Explanation:
Given;
mass of the man, m = 92 kg
displacement of the car's spring, x = 4.5 cm = 0.045 m
acceleration due to gravity, g = 9.8 m/s²
The spring constant of the car,
f = kx
where;
f is the weight of the man on the car = mg
mg = kx
k = mg/x
k = (92 x 9.8) / 0.045
k = 20,035.56 N/m
The angular speed of car, ω, when the is inside is given as 4.52 rad/s
The total mass of the car and the man is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\m = \frac{k}{\omega^2} = \frac{20,035.56}{(4.52)^2} = 980.7 \ kg[/tex]
The mass of the car alone = 980.7 kg - 92 kg
= 888.7 kg
≅ 890 kg
Therefore, the mass of the car is 890 kg
CR Physical Science B (GP) 20-21 / 1 Motion
1. A distance-time graph indicates that an object travels 2 m in 2 s and then travels another 80 m during the next 40 s.
What is the average speed of the object?
O 4 m/s
O
2 m/s
O 8 m/s
10 m/s
Answer:
Average speed = total distance traveled / total time traveled
V = (2 + 80) / (2 + 40) = 82 / 42 = 2 m / s
A child is playing in a park on a rotating cylinder of radius, r , is set in rotation at an angular speed of w. The Base of the cylinder is slowly moved away, leasing the child suspended against the wall in a vertical position.
What Is the minimum coefficient of friction between the child's clothing and wall is needed to prevent it from falling .
Answer:
[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]
Explanation:
From the given information:
The force applied to the child should be at equilibrium in order to maintain him vertically hung on the wall.
Also, the frictional force acting on the child against gravitational pull is:
[tex]F_f = \mu _sN[/tex]
where,
the centripetal force [tex]F_c[/tex] acting outward on the child is equal to the normal force.
[tex]F_c= N[/tex]
SO,
[tex]F_f = \mu_s F_c[/tex]
Since the centripetal force [tex]F_c = \dfrac{mv^2}{r}[/tex]
Then:
[tex]F_f = \dfrac{ \mu_s \times mv^2}{r}[/tex]
Using Newton's law, the frictional force must be equal to the weight
[tex]F_f = W[/tex]
[tex]\dfrac{ \mu_s \times mv^2}{r} = mg[/tex]
[tex]\dfrac{ \mu_s v^2}{r} = g[/tex]
Recall that:
The angular speed [tex]\omega = \dfrac{v}{r}[/tex]
Therefore;
[tex]g = \mu_s \omega^2 r[/tex]
Making the coefficient of friction [tex]\mu_s[/tex] the subject of the formula:
[tex]\mathbf{\mu_s = \dfrac{g}{\omega^2r}}[/tex]
An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surface, in terms of g, the acceleration of gravity at Earth's surface. A planet's gravitational acceleration is given by gp = G Mp/r^2p
a. 12.0 g.
b. 48.0 g.
c. 6.00 g.
d. 96.0 g.
e. 24.0 g.
Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
what is the main limitation of debye huckel theory
Answer:
Explanation:
For very low values of the ionic strength the value of the denominator in the expression above becomes nearly equal to one. In this situation the mean activity coefficient is proportional to the square root of the ionic strength. This is known as the Debye–Hückel limiting law.
Which electromagnetic waves have the greatest frequencies?
Answer:
Gamma rays
Explanation:
Gamma rays have the highest frequency in the electro magnetic spectrum
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
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When a Frisbee is flying horizontally through the air, the speed of the air flowing over the Frisbee's top is
Group of answer choices.
A. higher than the speed of the air beneath it and the pressure on top of the Frisbee is greater than the pressure beneath it.
B. lower than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.
C. lower than the speed of the air beneath it and the pressure on top of the Frisbee is greater than the pressure beneath it.
D. higher than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.
Answer:
D. higher than the speed of the air beneath it and the pressure on top of the Frisbee is less than the pressure beneath it.
Explanation:
Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.
Mathematically, it is given by the formula;
[tex] Acceleration = \frac {Net \; force}{mass} [/tex]
When a frisbee flies horizontally through the air, it experiences air above and beneath it. Also, there is a pressure acting on both on top and below the frisbee.
Typically, the air flowing over the top of the frisbee has a speed that is higher than the speed of the air beneath it.
Furthermore, the frisbee is able to remain in flight because the pressure on top of the frisbee is less than the pressure beneath it.
Lực tương tác giữa hai điện tích điểm khi đặt trong không khí là 1,5 N. Nhúng hai điện tích đó vào môi trường điện môi có hằng số điện môi là 3 thì lực tương tác giữa chúng là bao nhiêu?
Answer:
The force is now 0.5 N.
Explanation:
Force = 1.5 N
dielectric constant , k = 3
Let the two charges are q and q' and the distance between them is r.
The electrostatic force between the two charges is given by
[tex]F \alpha \frac{ q q'}{r^2}..... (1)[/tex]
When a dielectric material is inserted between the two charges, the new force is
[tex]F' \alpha \frac{ q q'}{kr^2}..... (2)[/tex]
From (1) and (2)
F' = F/K = 1.5/3 = 0.5 N
What is the change in internal energy if 70 J of heat is added to a system and
the system does 30 J of work on the surroundings. Uze al-Q-W.
O A. 40 J
O B. -40.3
O C. 100.
D. -1003
Answer:
A. 40 J
Explanation:
Given;
heat added to the system, Q = 70 J
work done by the system, W = 30 J
The change in the internal energy of the system is calculated using the first law of thermodynamic as shown below;
ΔU = Q - W
ΔU = 70 J - 30 J
ΔU = 40 J
Therefore, the change in the internal energy of the system is 40J
The slope at point A of the graph given below is:
WILL MARK BRAINLIEST TO CORRECT ANSWER
RQ/PQ I think
rise/run
Diwn unscramble the word
Answer:
WIND Is what you're looking for
Explanation:
The word is WIND
High speed stroboscopic photographs show that the head of a 183 g golf club is traveling at 58.6 m/s just before it strikes a 46.6 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.
Answer:
The speed of the golf ball just after the impact is 73.04 m/s.
Explanation:
Given that,
The mass of golf club, m₁ = 183 g = 0.183 kg
The mass of golf ball, m₂ = 46.6 g = 0.0466 kg
The initial speed of golf club, u₁ = 58.6 m/s
The initial speed of a golf ball, u₂ = 0
The final speeds of club, v₁ = 40 m/s
We need to find the speed of the golf ball just after impact. Using the conservation of momentum to find it.
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\m_1u_1=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.183 (58.6)-0.183(40)}{0.0466 }\\\\=73.04\ m/s[/tex]
So, the speed of the golf ball just after the impact is 73.04 m/s.
cho hệ cơ học như hình vẽ hai đầu dây buộc hai vật có khối lượng tương ứng là m1=2kg và m2>m1 lấy g=10m/s sau 1s kể từ lúc bắt đầu chuyển dộng hệ vật đi được 50 cm tính m2 và sức căng của dây
xin lỗi không có sơ đồ vui lòng cho biết sơ đồ

Two astronauts, each having a mass of 88.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.40 m/s. Treating the astronauts as particles, calculate each of the following.
a. the magnitude of the angular momentum of the system
b. the rotational energy of the system
c. What is the new angular momentum of the system?
d. What are their new speeds?
e. What is the new rotational energy of the system
Answer:
a) L = 4.75 103 kg m² / s, b) K_total = 2.57 10³ J,
c) L₀ = L_f =4.75 103 kg m² / s, d) K = 1.03 10⁴ J, K = 1.03 10⁴ J
Explanation:
a) the angular momentum is the sum of the angular momentum of each astronaut
the distance is measured from the center of the circle r = 10/2 = 5.0 m
L = 2m v r
L = 2 88.0 5.40 5.0
L = 4.75 103 kg m² / s
b) rotational kinetic energy
K = ½ I w²
As there are two astronauts, the total energy is the sum of the energy of each no.
The moment of inertia of a point mass
I = m r²
I = 88 5²
I = 2.2 10³ kg m²
the angular velocity is given by
v = w r
w = v / r
w = 5.40 / 5
w = 1.08 rad / s
the kinetic energy of the system
K_total = 2 K
K_total = 2 (½ I w²)
K_total = 2.2 10³ 1.08²
K_total = 2.57 10³ J
c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half
r = 2.5 m
I = 88 2.5²
I = 5.5 10² kg m²
for the change in angular velocity let us use the conservation of moment
L₀ = L_f
2Io wo = 2 I w
w = Io / I wo
w = 2.2 10³ / 5.5 10² 1.08
w = 4.32 rad / s
linear velocity is
v = w r
v = 4.32 2.5
K = 1.03 10⁴ J
the kinetic energy of the system is
K = 5.5 10² 4.32²
K = 1.03 10⁴ J
What is the mass of an object that experiences a gravitational force of 510 N near Earth's surface?
53.0 kg
52.0 kg
51.0 kg
54.1 kg
Answer:
52.006 Kilograms
.............................................
The mass of an object that experience a gravitional force of 510 N near earths surface in 52.0 kg
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
A pair of butterflies reproduces and has one thousand offspring. All one thousand of the offspring have the alleles Aa. What is the most likely combination of alleles (genotype) for each parent?
Answer:
Alleles AA and aa
Explanation:
For all the offsprings of the butterflies to have the same heterozygous alleles Aa, it means the two parents have different homogyzous alleles. That is, one of the parents had the alleles AA while the other had the alleles aa. Thus, a combination of the two pairs of alleles will produce 100% Aa alleles in the offspring as seen in the image attached.
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
A power plant generates 150 MW of electrical power. It uses a supply of 1000 MW from a geothermal source and rejects energy to the atmosphere. Find the power to the air and how much air should be flowed to the cooling tower (kg/s) if its temperature cannot be increased more than 10oC.
Answer:
- the power to the air is 850 MW
- mass flow rate of the air is 84577.11 kg/s
Explanation:
Given the data in the question;
Net power generated; [tex]W_{net[/tex] = 150 MW
Heat input; [tex]Q_k[/tex] = 1000 MW
Power to air = ?
For closed cycles
Power to air Q₀ = Heat input; [tex]Q_k[/tex] - Net power generated; [tex]W_{net[/tex]
we substitute
Power to air Q₀ = 1000 - 150
Q₀ = 850 MW
Therefore, the power to the air is 850 MW
given that ΔT = 10 °C
mass flow rate of air required will be;
⇒ Q₀ / CpΔT
we know that specific heat of air at p=c ; Cp = 1.005 kJ/kg.K
we substitute
⇒ ( 850 × 10³ ) / [ 1.005 × 10 ]
⇒ ( 850 × 10³ ) / 10.05
⇒ 84577.11 kg/s
Therefore, mass flow rate of the air is 84577.11 kg/s
A 2 kg stone is dropped from a height of 100 m. How far does it travel in the third second? take g = 9.8 m/s2
Answer:
S = 1/2 gt² = 1/2 × 9.8 × 3² = 4.9×9 = 44.1 m
Explanation:
Which of the following statements is false?
Weight is a vector quantity
Weight is measured in newtons. N
The weight of an object is the same on the Earth and the moon
Answer:
the weight of an object is the same on earth and moon
Explanation:
bcoz weight depends on both mass and gravity
since the gravity of earth and moon is different then the weight is also different
mass doesn't change not weight
