1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
juin [17]
3 years ago
11

show answer No Attempt Treated as a projectile, what is the maximum range, in meters, obtainable by a person if he or she has a

take-off speed of 9.2 m/s? Assume the motion is over level ground and the initial velocity makes an angle of 45° with the horizontal.

Physics
2 answers:
Lubov Fominskaja [6]3 years ago
7 0

Answer:

8.637 m

Explanation:

Using

R = (U²sin2θ)/g.................. Equation 1

Where R = Range, U = initial velocity, θ = angle of projection, g = acceleration due to gravity.

Note: For maximum Range, θ = 45°, then sin2θ = sin90° = 1

Therefore,

Rmax = U²/g..................... Equation 2

Given: U = 9.2 m/s, g = 9.8 m/s²

Substitute into equation 2

Rmax = 9.2²/9.8

Rmax = 84.64/9.8

Rmax = 8.637 m

Hence the maximum range = 8.637 m

Katen [24]3 years ago
3 0

Answer: R=8.64 meters

Explanation:

in the attachment

You might be interested in
A 300-kg bear grasping a vertical tree slides down at constant velocity. the friction force between the tree and the bear is:___
Nady [450]

3000 N of friction exists between the bear and the tree.

The bear goes down at a constant speed, thus there is no acceleration. The forces add up to zero. As a result, upward friction and downward weight result in zero  

i.e.,

F-mg=0

F = mg

F = (300 kg)/10 m/s^{2} (gravitational acceleration)

3000 N frictional force equals F.

How does friction force work?

  • Friction is the force that prevents two solid objects from rolling or sliding over one another. Even though frictional forces, such as the traction required to walk without slipping, may be advantageous, they can also be a significant hindrance to motion.

Between solid surfaces, there are three basic types of friction:

  • Rolling
  • Sliding.
  • Static.                                                                                                                   They are graded from strongest to weakest. Fluid friction happens when liquids or gasses are mixed together.

To learn more about Friction force, visit:

brainly.com/question/13707283

#SPJ4

3 0
2 years ago
How much work, in N*m, is done when a 10.0 N force moves an object 2.5 m?
Alik [6]
W = F * d
W = 10N * 2.5 m
W = 25 N m
So the answer you want is the third one down.
8 0
3 years ago
The light reactions could be viewed as analogous to a hydro-electric dam. In that case, the wall of the dam that holds back the
Viktor [21]

The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.

In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.

If you need to learn more about light reactions click here:

brainly.com/question/26623807

#SPJ4

3 0
1 year ago
The pressure on a balloon holding 356 mL of an ideal gas is increased from 267 torr to 1.00 atm. What is the new volume of the b
Natalija [7]

Answer:

124.52 mL

Explanation:

from Boyle's Law,

PV = P'V' ................... Equation 1

Where P = Initial pressure of the gas, V = Initial volume of the gas, P' = Final pressure of the gas, V' = Final volume of the gas.

make V' the subject of the equation.

V' = PV/P'............. Equation 2

Given: P = 267 torr = (267×0.00131) = 0.34977 atm, V = 356 mL, P' = 1 atm

Substitute into equation 2

V' = (0.34977×356)/1

V' = 124.52 mL.

Hence the new volume of the balloon = 124.52 mL

4 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
Other questions:
  • A magnesium surface has a work function of 3.60 eV. Electromagnetic waves with a wavelength of 320 nm strike the surface and eje
    14·1 answer
  • Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. spec
    15·1 answer
  • An electric motor and a single-fixed pulley work together to lift a 500 kg crate 50.0 m. How much work was done?
    8·2 answers
  • In a front-end collision, a 1500 kg car with shock-absorbing bumpers can withstand a maximumforce of 80 000 N before damage occu
    8·1 answer
  • (8th grade HELP)
    7·1 answer
  • You might have noticed that a feather falls slowly toward the ground, while a ball falls rapidly. Which statement correctly expl
    11·1 answer
  • 3.Three resistors of 25.0Ω, 30.0Ω, and 40.0Ω are in a series circuit with a 6.0-volt battery. What is the current in the circuit
    15·1 answer
  • Please help! i'll give brainliest!!
    5·2 answers
  • Help thank you. will mark as brainliest if correct and no link to some unknown websites
    5·2 answers
  • photon strikes the surface of tungsten and an electron is emitted. what is the maximum possible speed of the electron?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!