when a car suddenly stops,the objects in the backseat are thrown forward.this is explained by-
A.Newton's first law of motion
B.Newton's second law of motion
C.Newton's third law of motion
D.the universal law of gravitation
Answer:law of interia(1st law)
Explanation:
A man travels 25 kilometers in 15 minutes. A police officer stops him and
writes a ticket for speeding. The posted speed limit is 80 kilometers per
hour. Did the man exceed the posted speed limit?
Answer:
yes
Explanation:
cuz 15x2=30 30x2=60 60=1 hour, and the limited is 80 km per hour
15=25
15x4=60 25x4=100
so yes
he was going 100km per hour
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 miles at 20° W of S. Next, she rides 1.6 miles at 30° S of W. And finally, she rides 2.6 miles at 15° W of N. What is her final displacement from her starting point?
Answer:
D = 1.8677 miles , θ = 24.28º at South of West
Explanation:
This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry
first displacement. d = 1.2 miles to 30º south of East
cos ( 360-30) = cos (-30) = x₁ / d
sin (-30) = y₁ / d
x₁ = d cos (-30)
y₁ = d sin (-30)
x₁ = 1.2 cos (-30) = 1,039 miles
y₁ = 1.2 sin (-30) = -0.6 miles
second shift. d = 2.0 miles to 20º West of South
cos (270-20) = x₂ / d
cos (250) = y₂ / d
x₂ = 2.0 cos 250 = -0.684 miles
y₂ = 2.0 sin250 = -1.879 miles
Third displacement. d = 1.6 miles to 30º South of West
cos (180 + 30) = x₃ / d
sin (210) = y₃ / d
x₃ = 1.6 cos 210 = -1.3856 miles
y₃ = 1.6 sin 210 = -0.8 miles
Fourth displacement. d = 2.6 miles to 15º West of North
cos (90 + 15) = x₄ / d
sin (105) = y₄ / d
x₄ = 2.6 cos 105 = -0.6729 miles
y₄ = 2.6 sin 105 = 2,511 miles
having all the components we add
x-axis (West-East direction)
X = x₁ + x₂ + x₃ + x₄
X = 1.039 -0.684 - 1.3846 - 0.6729
X = -1.7025 miles
Y = y₁ + y₂ + y₃ + y₄
Y = -0.6 -1.879 -0.8 +2.511
Y = -0.768
The modulus of this displacement is we use the Pythagorean theorem
D = √ (X² + Y²)
D = √ (1.7025² + 0.768²)
D = 1.8677 miles
let's use trigonometry to find the direction
tan θ = Y / X
θ = tan⁻¹ Y / x
θ = tan⁻¹ (0.768 / 1.7025)
θ = 24.28º
as the two components are negative this angle is in the third quadrant
therefore in cardinal direction form is
θ = 24.28º at South of West
When dribbling you should?
a. Use both feet.
b.Use frequent small touches of the ball.
C.Protect the ball with your body.
D. All of the above.
A stunt man launches a car from the edge of a cliff horizontally. If he is launching off a
30-meter high cliff and must hit a mark 268-meters down range, with what speed
must he leave the cliff?
Answer:
All i got to say is that he is dead dead
Explanation:
Which combination of a wire coil and a core would make the weakest
electromagnet?
O A. 20 coils of bare wire around a steel core
B. 20 coils of insulated wire around a hollow cardboard core
O C. 50 coils of insulated wire around a hollow cardboard core
D. 50 coils of bare wire around a steel bolt
The combination of a wire coil and a core would make the weakest electromagnet 20 coils of insulated wire around a hollow cardboard core. Thus, option B is correct.
Electromagnet is a type of magnet that produces its magnetic field when an electric field is applied across the material. The electromagnet is the temporary magnet as it acquires the magnetic properties when the electric current passes through it. When there is no electric current, its magnetic properties disappear.
Electromagnet consists of a wire wound on the coil and the current passes through the wire. The wire wound on the coil is mostly a ferromagnetic material like iron and the coil is a soft iron core. In an electromagnet, the magnetic field depends on the number of turns of the coil and the current flows through it.
From the given, steel core is wounded with wire it makes a strong electromagnet depending on the number of turns of the coil. The cardboard core is not a ferromagnetic material and hence, a small number of turns over the cardboard forms the weakest electromagnet.
Hence, the ideal solution is option B)20 coils of insulated wire around a hollow cardboard core.
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What is the speed of a
rocket that travels
8,000m in 13 seconds?
Answer:
speed of rocket=615.38m/s
Explanation:
Which of the following statements is true?
Observations are often used to form questions about the world.
The dependent variable is changed by the experimenter.
The best hypotheses are written as answers to a question.
A hypothesis is created at the end of an experiment.
Answer:
The answer is B.
Explanation:
They are in control of the experiment, they can change it the variables to better help the experiment.
Answer:
b
Explanation: its b
2. What is the accelaration of a
4000 kg airplane that has a force
of 24,000 N?
Answer:
c) 6 m/s²
Explanation:
A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed for the ball to reach its max height.
Answer:
B) t = 1.83 [s]
A) y = 16.51 [m]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f} =v_{o} -g*t[/tex]
where:
Vf = final velocity = 0
Vo = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81 [m/s²]
t = time [s]
Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.
A) The maximum height is reached when the final velocity of the ball is zero.
0 = 18 - (9.81*t)
9.81*t = 18
t = 18/9.81
t = 1.83 [s], we found the answer for B.
Now using the following equation.
[tex]y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\[/tex]
where:
y = elevation [m]
Yo = initial elevation = 0
y = 18*(1.83) - 0.5*9.81*(1.83)²
y = 16.51 [m]
A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete is = 0.02. (a) What is the cart's acceleration if rolling friction is neglected? (b) What is the cart's acceleration if rolling friction is included?
Answer:
(a) 5.62 m/s²
(b) 5.46 m/s²
Explanation:
The given values are:
mass,
m = 50 kg
angle
= 35°
(a)
If friction neglected,
⇒ [tex]F_x=mgSin \theta=ma[/tex]
⇒ [tex]a_x=gSin \theta[/tex]
[tex]=9.8 \ Sin35^{\circ}[/tex]
[tex]=5.62 \ m/s^2[/tex]
(b)
If friction present,
⇒ [tex]F_x=mgCos \theta[/tex]
⇒ [tex]F_x=mgSin \theta-\mu_r mgCos \theta[/tex]
⇒ [tex]a=gSin \theta-\mu_rgCos \theta[/tex]
[tex]=9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}[/tex]
[tex]=5.46 \ m/s^2[/tex]
A person travels by car from one city to another. He drives for 30.0 min at 80.0 km/h, 12.0 min at 105 km/h, and 45.0 min at 40.0 km/h, and he spends 15.0 min eating lunch and buying gas. What is the average speed for the trip?
Answer:
53.53 km/h
Explanation:
Given that the person drives for 30.0 min at 80.0 km/h, 12.0 min at 105 km/h, and 45.0 min at 40.0 km/h, and he spends 15.0 min eating lunch and buying gas.
So, the total time to complete the journey is
t=30+12+45+15=102 min
[tex]=\frac{102}{60}=1.7[/tex] hours
As distance=(speed)x(time), so
the distance covered in 30 minutes with a speed of 80 km/h
[tex]d_1=80 \times \frac{30}{60}=40[/tex] km.
The distance covered in 12 minutes with a speed of 105 km/h
[tex]d_2=105 \times \frac{12}{60}=21 km[/tex].
The distance covered in 45 minutes with a speed of 40 km/h
[tex]d_3=40 \times \frac{45}{60}=30 km[/tex].
So, to the total distance covered is
[tex]d=d_1+d_2+d_3[/tex]
[tex]=40+21+30=91 km[/tex]
As the average speed = (Total distance covered)/(Total time taken)
So, the average speed for the whole journey
[tex]=\frac{d}{t}[/tex]
[tex]=\frac{91}{1.7}= 53.53 km/h[/tex].
Hence, the average speed for the trip is 53.53 km/h.
How do you draw a free-body diagram of an object that is attached to a string moving in uniform circular motion? What forces do you draw?
Answer:Whenever an object experiences uniform circular motion there will always be a net force acting on the object pointing towards the center of the circular path. This net force has the special form , and because it points in to the center of the circle, at right angles to the velocity, the force will change the direction of the velocity but not the magnitude.
It's useful to look at some examples to see how we deal with situations involving uniform circular motion.
Example 1 - Twirling an object tied to a rope in a horizontal circle. (Note that the object travels in a horizontal circle, but the rope itself is not horizontal). If the tension in the rope is 100 N, the object's mass is 3.7 kg, and the rope is 1.4 m long, what is the angle of the rope with respect to the horizontal, and what is the speed of the object?
As always, the place to start is with a free-body diagram, which just has two forces, the tension and the weight. It's simplest to choose a coordinate system that is horizontal and vertical, because the centripetal acceleration will be horizontal, and there is no vertical acceleration.
The tension, T, gets split into horizontal and vertical components. We don't know the angle, but that's OK because we can solve for it. Adding forces in the y direction gives:
This can be solved to get the angle:
In the x direction there's just the one force, the horizontal component of the tension, which we'll set equal to the mass times the centripetal acceleration:
We know mass and tension and the angle, but we have to be careful with r, because it is not simply the length of the rope. It is the horizontal component of the 1.4 m (let's call this L, for length), so there's a factor of the cosine coming in to the r as well.
Rearranging this to solve for the speed gives:
which gives a speed of v = 5.73 m/s.
Example 2 - Identical objects on a turntable, different distances from the center. Let's not worry about doing a full analysis with numbers; instead, let's draw the free-body diagram, and then see if we can understand why the outer objects get thrown off the turntable at a lower rotational speed than objects closer to the center.
In this case, the free-body diagram has three forces, the force of gravity, the normal force, and a frictional force. The friction here is static friction, because even though the objects are moving, they are not moving relative to the turntable. If there is no relative motion, you have static friction. The frictional force also points towards the center; the frictional force acts to oppose any relative motion, and the object has a tendency to go in a straight line which, relative to the turntable, would carry it away from the center. So, a static frictional force points in towards the center.
Summing forces in the y-direction tells us that the normal force is equal in magnitude to the weight. In the x-direction, the only force there is is the frictional force.
The maximum possible value of the static force of friction is
As the velocity increases, the frictional force has to increase to provide the necessary force required to keep the object spinning in a circle. If we continue to increase the rotation rate of the turntable, thereby increasing the speed of an object sitting on it, at some point the frictional force won't be large enough to keep the object traveling in a circle, and the object will move towards the outside of the turntable and fall off.
Why does this happen to the outer objects first? Because the speed they're going is proportional to the radius (v = circumference / period), so the frictional force necessary to keep an object spinning on the turntable ends up also being proportional to the radius. More force is needed for the outer objects at a given rotation rate, and they'll reach the maximum frictional force limit before the inner objects will.
Explanation:
does anyone know how to do this? the scenario is ""dominique reads that race cars have wide tires because the increased area of contact between the tire and the road results in a stronger force of friction. she hypothesizes that the force of kinetic friction on an object is directly proportional to the area of the object in contact with the surface and wants to test this hypothesis. Dominique and Blake take a long wooden plank and cut the plank into pieces that have different lengths but the same width and height. The students also have access to other equipment commonly available in a school physics lab.
Answer:
Stupid
Explanation:
Because there is never a answer when we are trying to find one
A football is thrown at an angle of 30° above the ground and is in the air for a total of 1.93 s
before it is caught (at the same height). If the football traveled a horizontal distance of 43 m
a) with what speed was the football thrown?
b) what max height would the football reach?
1.53 secs :)))))))))))))))
Athlete’s foot is also called...
A.) pedicure sickness Not Quite!
B.) runner’s bacteria Not Quite!
C.) toe virus Not Quite!
D.) tinea pedis
Answer:
Your answer is: D) Tinea Pedis
Explanation:
Hope this helped : )
Beginning: Identify the above circuits (A and B) as either a series or parallel circuit.
The diagram below shows the velocities of two runners.
11 m/s west
Runner 1
5 m/s east
Runner 2
From the frame of reference of runner 1, what is the velocity of runner 2?
O A. 5 m/s east
O B. 6 m/s east
C. 16 m/s west
O D. 16 m/s east
Answer : 16 m/s east
Explanation: Trust the process!!
From the frame of reference of runner 1, the velocity of runner 2 will be D) 16 m/s east
What is relative velocity ?
The relative velocity is defined as the velocity of an object with respect to another observer
When two bodies are moving in opposite directions , relative velocity get added and if bodies are moving in same direction , relative velocity will get subtracted .
From the frame of reference of runner 1, the velocity of runner 2 will be
v (relative) = v1 + v1
v1 = velocity of runner 1 = 11m/s west
v2 = velocity of runner 2 = 5m/s
v = 11 + 5 = 16 m/s
correct answer will be O D. 16 m/s east as runner 2 is running in east direction
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The lens in the eyepiece of a reflecting telescope breaks. How will this most affect the function of the telescope?
A. Magnified images will appear upside down. B. Magnified images will not be created.
C. Light from distant objects will not be captured.
D. Light rays will not be sent to the eyepiece.
Answer:
B
Magnified images will not be created.
Explanation:
I did it and this was the correct answer
Answer:
B is correct the person above me is also correct. Hope this helps
Explanation:
Which is the balanced equation for Sg + O2 → SO2?
OS: + 016 — 8S02
O S₂ + O₂S₃ + O₂
O S8+ O₂S₂O₂
O Sg + 802 8S02
The balanced equation for S(g) + O₂ → SO₂ is as follows: S₈ + 8O₂ → 8SO₂ (option C).
What is a balanced equation?A balanced equation is a chemical equation in which the number of atoms of each element on both sides of the equation are the same.
This means that the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.
Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products.
According to this question, sulfur reacts with oxygen gas to produce sulfur dioxide as follows:
S₈ + O₂ → SO₂
The balanced chemical equation is as follows:
S₈ + 8O₂ → 8SO₂
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Los satélites de televisión giran alrededor de la Tierra en una órbita de 42370 km de radio. a) ¿Cuánto vale la aceleración de la gravedad en esa órbita? (Resultado: g = 0,22 m/s2 ) b) ¿Cuánto pesará allí un satélite de 1200 kg? (Resultado: P = 264 N) Datos: G = 6,67 10-11 Nm2 /kg masa de la Tierra: 6 1024 kg
Answer:
(a) La aceleración de la gravedad en esa órbita es 0.22 [tex]\frac{m}{s^{2} }[/tex]
(b) Un satélite de 1,200 kg pesará allí 264 N.
Explanation:
(a) Cada punto del espacio tiene un valor del campo proporcional a la función, siguiendo el modelo de gravitación de Newton:
[tex]g=G*\frac{m}{r^{2} }[/tex]
donde m será masa, r la distancia entre los objetos y G la constante de gravitación universal, cuyo valor es [tex]G=6.672*10^{-11}\frac{N*m^{2} }{kg }[/tex]
En este caso:
m= 6*10²⁴ kgr=42,370 km= 42,370,000 mReemplazando:
[tex]g=6.672*10^{-11}\frac{N*m^{2} }{kg }*\frac{6*10^{24}kg }{(42,370,000 m)^{2} }[/tex]
Resolviendo, obtenes:
g= 0.22 [tex]\frac{m}{s^{2} }[/tex]
La aceleración de la gravedad en esa órbita es 0.22 [tex]\frac{m}{s^{2} }[/tex]
(b) La ley fundamental de la dinámica o segunda ley de Newton expresa que la fuerza neta que es aplicada sobre un cuerpo es proporcional a la aceleración que adquiere en su trayectoria. Esto se expresa matemáticamente como:
F= m*a
en donde F es fuerza neta , m la masa y a la aceleración.
El peso es la fuerza que ejerce la gravedad sobre una masa, por lo que se obtiene de manera análoga a la fuerza:
P=m*g
En este caso la masa m tiene un valor de 1,200 kg y la gravedad g un valor de 0.22 [tex]\frac{m}{s^{2} }[/tex].
Reemplazando:
P=1,200 kg*0.22 [tex]\frac{m}{s^{2} }[/tex]
Resolviendo:
P= 264 N
Un satélite de 1,200 kg pesará allí 264 N.
While sitting in class, your body exerts a force of 600 N on your chair. How much work do you do?
Answer:
1000 N
Explanation:
What is the speed of a car that
accelerates at a rate of 10 m/s2 for a
total of 25 seconds?
Answer:
250 m/sExplanation:
The speed of the car can be found by using the formula
v = a × twhere
a is the acceleration
v is the speed
t is the time
From the question we have
v = 10 × 25
We have the final answer as
250 m/sHope this helps you
Marlene loves astronomy and has enrolled in an online course to learn more about space. The course gives Marlene access to a digital classroom.
What does this enable her to do? Check all that apply.
discuss the solar system with others in the class
attend a field trip to a planetarium at a local college
upload an amateur video of a shooting star
make a presentation about her favorite planet
take notes, tests, and quizzes
watch instructional videos made by a teacher
Answer:
A E and F are the correct answers
Explanation:
Answer:
A E F are correct
Explanation:
PLEASE HELP THE question on picture
Vertical:
(20 m/s) sin(25º) ≈ 8.45 m/s
Horizontal:
(20 m/s) cos(25º) ≈ 18.1 m/s
A dumptruck filled with sand moves 1.8 km/h when it begins to accelerate uniformly at a constant rate. After traveling 4.0 102 m, the truck's speed is 24.0 km/h. What is the magnitude of the truck's acceleration?
The acceleration of the truck is obtained from the calculation done as 0.055 m/s^2.
What is the magnitude of the truck's acceleration?We know that the acceleration is defined as the rate of the change of the velocity of the object with time and we can be able to obtain the acceleration by the use of the equations of kinematics as usual.
Initial velocity = 1.8 km/h or 0.5 m/s
Final velocity = 24.0 km/h or 6.67 m/s
Distance covered = 4.0 * 10^2 m
Then;
v^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
s = distance covered
a = v^2 - u^2/2s
a = (6.67)^2 - (0.5)^2/2 * 4.0 * 10^2
a = 44.4 - 0.25/800
a = 0.055 m/s^2
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hey pls help with this HDJSJDH im rlly bad at this kinda stuff :,) i'll mark the correct answer brainliest
Answer:
A. Cue, White Ball, Blue Ball
Explanation:
Visualize in which order things are being hit
HELP PLEASE
A load is a device that changes electrical energy into other forms of energy. Which load converts electrical energy into thermal energy by passing an electric current through a high-resistance material?
A.
Hand-powered generator
B.
Toaster
C.
Fan
D.
Electrical drill
Answer:
A toaster because you plug it up for electric energy and heat to toast it
A marble rolls off a table from a height of 1.5 meters. Approximately how much time was the marble in the air?
O 1.1 seconds
O 0.78 seconds
O 0.55 seconds
O 0.2 seconds
Answer:
t = 0.55 s
Explanation:
Given that,
Height of the marble, h = 1.5 m
We need to find the time for which the marble is in air. Let the time is t. Using second equation of motion to find it as follows :
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Here, u = 0 and a = g
[tex]h=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\cdot1.5}{9.8}}\\\\t=0.55\ s[/tex]
So, the marble is in air for 0.55 s. Hence, the correct option is (c).
find acceleration of block and tension in strings if m1 =8kg m2= 4kg and take g=10m/s²
[tex]\sf{m_1}[/tex] = 8kg
[tex]\sf{m_2}[/tex] = 4kg
g = 10m/s²
[tex]\large\bold{\underline{\underline{To \: Find:-}}}[/tex]→Acceleration of block
→Tension in strings
[tex]\large\bold{\underline{\underline{Solution:-}}}[/tex]Let, the two equation will be ,
[tex]\leadsto[/tex] 8g - T = 8a ..... Equation no (1)
[tex]\leadsto[/tex] T - 4g = 4a ..... Equation no (2)
Putting the value of equation no (1) and (2) we get,
[tex]\implies[/tex] 4g = 12a
[tex]\implies[/tex] 4 × 10 = 12a
[tex]\implies[/tex] 40 = 12a
[tex]\implies[/tex] a = [tex]\sf\dfrac{\cancel{40}}{\cancel{12}}[/tex]
[tex]\dashrightarrow[/tex] a = [tex]\dfrac{10}{3}[/tex]
Again, we have to putting the value of a in the equation no (2) we get,
[tex]\implies[/tex] T - 4g = 4 × [tex]\dfrac{10}{3}[/tex]
[tex]\implies[/tex] T - 4 × 10 = 4 × [tex]\dfrac{10}{3}[/tex]
[tex]\implies[/tex] T - 40 = [tex]\dfrac{40}{3}[/tex]
[tex]\implies[/tex] T = [tex]\dfrac{40}{3}[/tex] + 40
[tex]\implies[/tex] T = [tex]\dfrac{40 + 120}{3}[/tex]
[tex]\dashrightarrow[/tex] T = [tex]\dfrac{160}{3}[/tex] N
[tex]\therefore[/tex] Acceleration of block = [tex]\dfrac{10}{3}[/tex] m/s².
And, Tension in strings = [tex]\dfrac{160}{3}[/tex] N
