Answer:
The speed of the flashlight at that point is 3.7 m/s
Explanation:
When an object of mass M is at a height H above the ground, the potential energy of the object is:
U = M*H*g
Where g is the gravitational acceleration, g = 9.8 m/s^2
And for an object with velocity v, the kinetic energy is:
K = (M/2)*v^2
We know that when the flashlight of mass 1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:
M = 1kg
H = 0.7m
g = 9.8 m/s^2
Replacing these in the equations, we get:
U = K
(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2
As the mass factor appears in both sides, we can remove it:
(0.7 m)*(9.8 m/s^2) = (v^2)/2
Now we can multiply both sides by 2:
2*(0.7 m)*(9.8 m/s^2) = v^2
Now let's apply the square root to both sides:
√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s
Look at the Position vs. Time and Velocity vs. Time plots. What is the person's velocity when his position is at its maximum value (around 6 m )
Answer:
The person's velocity is zero.
Explanation:
A 0.413 kg block requires 1.09 N
of force to overcome static
friction. What is the coefficient
of static friction?
(No unit)
PLEASE HELP!
Answer:
static friction=0.126
How are soil and air similar?
Answer:
The air in the soil is similar in composition to that in the atmosphere with the exception of oxygen, carbon dioxide, and water vapor. In soil air as in the atmosphere, nitrogen gas (dinitrogen) comprises about 78%. In the atmosphere, oxygen comprises about 21% and carbon dioxide comprises about 0.036%.
hope this helps
have a good day :)
Explanation:
Which of the following is form of energy:
a) Power
b) Light
C) pressure
d) None
Answer:
Explanation:
b) light
4- What force must be applied to a surface area of 0.0025m , to create a pressure ol
200.000Pa?
An irrigation canal has a rectangular cross section. At one point where the canal is 18.2 m wide and the water is 3.55 m deep, the water flows at 2.55 cm/s . At a second point downstream, but on the same level, the canal is 16.3 m wide, but the water flows at 11.6 cm/s . How deep is the water at this point
Answer:
Explanation:
Rate of volume flow at two points will be same at two points .
A₁ V₁ = A₂V₂
A₁ and A₂ are area of cross section at two points and V₁ and V₂ are velocities .
A₁ = 18.2 x 3.55 = 64.61 m²
V₁ = 2 .55 x 10⁻² m/s
A₂ = 16.3 x d = 16.3 d m²
d is depth at second point .
V₂ = 11.6 x 10⁻² m/s
64.61 m² x 2 .55 x 10⁻² m/s = 16.3 d m² x 11.6 x 10⁻² m/s
d = .87 m
so canal is .87 m deep.
. A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car.
four times
twice
half
equal to
Complete question is;
A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The
centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.
four times
twice
half
equal to
Answer:
Half
Explanation:
Formula for centripetal force is given as;
F = mv²/R
Where;
v is velocity
R is radius
Now, centripetal acceleration is given by;
a = v²/R
Since they both travel with the same velocity V and radius remains the same, we can say that;
F = ma
For the small car;
FS = ma
For the big car;
FL = 2ma
This means the force of the small car is half of that of the Large car
Thus;
FS = ½FL
A 35.0 g bullet strikes a 5.3 kg stationary wooden block and embeds itself in the block. The block and bullet fly off together at 7.1 m/s. What was the original speed of the bullet? (WILL GIVE BRAINLIEST)
Answer:
= 1200m/s or 1.2 x [tex]10^{3}[/tex] m/s
Explanation:
BRAINLIST A wave travels at a constant speed. How does the wavelength change if the
frequency is reduced by a factor of 3? Assume the speed of the wave remains
unchanged.
A. The wavelength does not change.
B. The wavelength increases by a factor of 3.
C. The wavelength decreases by a factor of 3.
D. The wavelength increases by a factor of 9.
There are three 20.0 Ω resistors connected in series across a 120 V generator
Answer:
That is equal to R1 + R2. If three or more unequal (or equal) resistors are connected in series then the equivalent resistance is: R1 + R2 + R3 +…, etc. One important point to remember about resistors in series networks to check that your maths is correct.
A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter. White light is incident normally on the grating. What is the longest wavelength that forms an intensity maximum in the fifth order
Answer:
377 nm
Explanation:
Number of lines per meter is, [tex]N &=530 \times 1000 \\ &=530000 \text { lines } / \mathrm{m} \end{aligned}[/tex]
Grating element is, [tex]d=\frac{1}{N}[/tex]
[tex]=1.8868 \times 10^{-6} \mathrm{~m}[tex]
Order is, n=5
Condition for maximum intensity is, [tex]d \sin \theta=n \lambda[/tex]
[tex]\lambda &=\frac{1.8868 \times 10^{-6}}{5(\sin 90)} \\ &=0.377 \times 10^{-6} \mathrm{~m} \\ &=377 \mathrm{~nm}[/tex]
24. A anvil with a mass of 60 kg falls from a height of 9.5 m. How fast is it going right
before it hits the ground?
V= I*R
V = voltage (measured in volts) V
I = current (measured in amperes) A
R = resistance (measured in Ohms) Ω
So they give us this
V=IR
V= 1.8
I=0.4
R=?
So we insert the thing that we know.
1.8=0.4*R
We need to leave our unknown value alone. So if our value of 0.4 is multiplying the unknown value it passes to the other side dividing.
So we have this.
Lastly we solve.
R=4.5ohms
The formula to find R is V=IR
V/I=R
So the resistance will be the Voltage divided by the Current
Help me with this please
Answer:
check out of phase
Explanation:
this is my answer
The door is 2 m tall. How tall is it in inches? Note: There are 2.54 cm in 1 inch.
A. 78.7 in
B. 500 in
C. 787.4 in
D. 201.4 in
Answer:
Height of the door = 2m = 2000 cm
1 in = 2.54 cm
So 1 cm = 1/2.54 in
2000 cm = 200000/ 254
=
787.401574803
So no.c is correct
The door is 78.7 inch tall. Hence, option (A) is correct.
What is unit of length?Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.
The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.
Given that: the height of the door is = 2 meter
= 2*100 centimeter
= 200 centimeter.
There are 2.54 centimeter in 1 inch.
Hence, the height of the door is = 2 meter = 200 centimeter
= (200/2.54) inch
= 78.7 inch.
The door is 78.7 inch tall.
Learn more about length here:
https://brainly.com/question/17139363
#SPJ2
5. Stopping a fast-moving object is harder than stopping a slow-moving
one.
True
False
3 نقطة (نقاط)
السؤال 2
A block slides on a rough horizontal surface from paint A to point B. A force (magnitude P -
2.0 N) acts on the block between A and B, as shown. If the kinetic energies of the block at A
and B are 5.0 J and 4.0 J, respectively and the work done on the block by the force of friction
as the block moves from A to B is -4.5 J what is the distance between point A and B?
40°
B
The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is
[tex]W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J[/tex]
Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.
The only other force acting on the block as it moves is the force P. Let [tex]W_P[/tex] be the work done by the force P. Then the total work done on the block is
[tex]W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}[/tex]
1. Pam has a mass of 48.3 kg and she is at rest on
smooth, level, frictionless ice. Pam straps on
a rocket pack. The rocket supplies a constant
force for 27.3 m and Pam acquires a speed of
62 m/s.
What is the magnitude of the force?
Answer in units of N.
2. What is Pam’s final kinetic energy?
Answer in units of J.
3. A child and sled with a combined mass of 55.7
kg slide down a frictionless hill that is 11.3 m
high at an angle of 29 ◦
from horizontal.
The acceleration of gravity is 9.81 m/s
3. If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s
Answer:
1. F = 3400 N = 3.4 KN
2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3. v = 14.9 m/s
Explanation:
1.
First, we will calculate the acceleration of Pam by using the third equation of motion:
[tex]2as = v_f^2-v_i^2[/tex]
where,
a = acceleration = ?
s = distance = 27.3 m
vf = final speed = 62 m/s
vi = initial speed = 0 m/s
Therefore,
[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]
Now, we will calculate the force by using Newton's Second Law of Motion:
F = ma
F = (48.3 kg)(70.4 m/s²)
F = 3400 N = 3.4 KN
2.
Final kinetic energy is given as:
[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]
[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]
3.
According to the law of conservation of energy:
[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]
where,
v = speed at bottom = ?
g = acceleration due to gravity = 9.81 m/s²
h = height at top = 11.3 m
Therefore,
[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]
v = 14.9 m/s
Moving current has electrical energy.
An athlete is performing squats in the weight room. The knee is going from anatomical position to 92 degrees and then back to anatomical position each squat. The athlete performs a total of 10 squats. This is done over a time period of 30 seconds. What is the angular acceleration (rad/sec 2) of the knee
Answer:
α = 0.357 ras / s²
Explanation:
This is a rotational kinematics exercise, it tells us that it performs 10 squats in 30 s, for which it performs one squat at t = 3 s, also indicates that the angle of the squat is θ = 92º
θ = θ₀ + w₀ t + ½ α t²
the athlete starts from rest, whereby w₀ = 0 and the initial angle in the vertical position is zero (θ₀=0)
θ = ½ α t²
α = 2 θ /t²
let's reduce the magnitudes to the SI system
θ = 92º (π rad /180º) = 0.511π rad
let's calculate
α = 2 0.5111π /3²
α = 0.1136π rad / s²
α = 0.357 ras / s²
g Monochromatic light with wavelength 633 nn passes through a narrow slit and a patternappears on a screen 6.0 m away. The distance on the screen between the centers of thefirst minima on either side of the screen is 32 mm. How wide (in mm) is the slit
Answer:
d = 1.19 x 10⁻⁴ m = 0.119 mm
Explanation:
This problem can be solved by using Young's double-slit experiment formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = fringe spacing = 32 mm = 0.032 m
L = slit to screen distance = 6 m
λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m
d = slit width = ?
Therefore,
[tex]0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}[/tex]
d = 1.19 x 10⁻⁴ m = 0.119 mm
A man on the Moon observes two spaceships coming toward him from opposite directions at speeds of 0.600c and 0.600c. What is the relative speed of the two ships as measured by a passenger on either one of the spaceships
Answer:
If we use the equation for the transformation of velocities for moving frames:
v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'
v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)
or v' = 1.2 c / (1 + .36) = .88 c
v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u (-.6 c) and we measure the speed of v as seen in the frame moving to the left
A retired bank president can easily read the fine print of the financial page when the newspaper is held no closer than arm's length, 59.1 cm from the eye. What should be the focal length of an eyeglass lens that will allow her to read at the more comfortable distance of
Answer:
Explanation:
comfortable distance is 25 cm .
He must be using convex lens . In that case rays coming from object placed at 25 cm appears to be coming from 59.1 cm due to converging nature of convex lens.
object distance u = -25 cm
image distance v = -59.1 cm
Lens formula
1 / v - 1 /u = 1 /f
-1 / 59.1 + 1 / 25 = 1/f
- .0169 + .04 = 1 / f
.0231 = 1 / f
f = 43.3 m
An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose the electron is released from rest a distance 0.050 m from the ring center. It then oscillates through the ring center. Calculate its period. (The electron is always much closer to the ring center than a radius.)
Answer:
T = 1.12 10⁻⁷ s
Explanation:
This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.
All the charge dq is at a distance r
dE = k dq / r²
Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry
cos θ =[tex]\frac{dE_x}{dE}[/tex]
dEₓ = dE cos θ
cos θ = x / r
substituting
dEₓ = [tex]k \frac{dq}{r^2 } \ \frac{x}{r}[/tex]
DEₓ = k dq x / r³
let's use the Pythagorean theorem to find the distance r
r² = x² + a²
where a is the radius of the ring
we substitute
dEₓ = [tex]k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq[/tex]
we integrate
∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} } ∫ dq
Eₓ = [tex]k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}[/tex]
In the exercise indicate that the electron is very central to the center of the ring
x << a
Eₓ = [tex]k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}[/tex]
if we expand in a series
[tex](\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2[/tex]
we keep the first term if x<<a
Eₓ = [tex]\frac{ k Q}{a^3} \ x[/tex]
the force is
F = q E
F = [tex]- \frac{kQ }{a^3} \ x[/tex]
this is a restoring force proportional to the displacement so the movement is simple harmonic,
F = m a
[tex]- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }[/tex]
[tex]\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x[/tex]
the solution is of type
x = A cos (wt + Ф)
with angular velocity
w² = [tex]\frac{keQ}{m a^3}[/tex]
angular velocity and period are related
w = 2π/ T
we substitute
4π² / T² = \frac{keQ}{m a^3}
T = 2π [tex]\sqrt{\frac{m a^3 }{keQ} }[/tex]
let's calculate
T = 2π [tex]\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }[/tex]
T = 2π pi [tex]\sqrt{320.426 \ 10^{-18} }[/tex]
T = 2π 17.9 10⁻⁹ s
T = 1.12 10⁻⁷ s
how dose an exam question outed from text book
Answer:
In which school you are???
Explanation:
If an electromagnetic wave has a frequency of 6×10^5 hz, what is its wavelength? what is its wavelength? A. 2 x 10^12m, B. 5 x 10^14m, C. 5 x 10^2m, 2 x 10^-3m
Answer:
5*10^2
Explanation:
A p e x
An old fashioned string of 80 Christmas lights is wired in series. Each bulb has a resistance of 2 Ohms and the entire string is plugged into a 120V outlet. What is the current passing through each of the bulbs?
The sum of the resistance = 2 ohms x 80 lights = 160 ohms.
Current = Total voltage / total resistance:
Current = 120V / 160 ohms
Current = 0.75 Amps
Can someone help me
paano matutugunan o matutulungan ng pamahalaan at ng mga guro yubg mga estudyanteng nakararanas nag stress at anxiety.
Answer: how the government and teachers can address or help students experiencing stress and anxiety.
Explanation:
What effect does the Duck Velocity have on the waves seen by the observer?
Towards the boat:
Away from the boat:
Same as the boat:
Why is it important for equipment for sport to be strong? To protect us
Answer:
To protect us.
Explanation:
For ex. your dunking on a basketball hoop if that wasn't strong you would fall on your back and get injured.