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Vlad [161]
3 years ago
10

A hailstone traveling with a velocity of 43 meters/second comes to a virtual stop 0.28 seconds after hitting water. What is the

magnitude of its acceleration in the water?
A. 12 meters/second2
B. 1.5 × 102 meters/second2
C. 78.6 meters/second2
D. 6.5 × 10-3 meters/second2
Physics
2 answers:
lisabon 2012 [21]3 years ago
5 0

Magnitude of acceleration = (change in speed) / (time for the change) .

Change in speed = (ending speed) - (starting speed)

                            =       zero            - (43 m/s)

                            =          -43 m/s .

Magnitude of acceleration = (-43 m/sec) / (0.28 sec)

                                          =  (-43 / 0.28)  (m/sec) / sec

                                          =        153.57...  m/s²

                                          =        1.5...  x 10²  m/s²  .

Andrei [34K]3 years ago
5 0

Answer:

it's b

Explanation:

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Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
How much heat is absorbed by a 450g gold block as energy from the sun causes its temperature to change from 27°C to 32°C? (Speci
kotegsom [21]

Answer:

Q = 1057.5 [cal]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

Q=m*C_{p}*(T_{final}-T_{initial})

where:

Q = heat energy [cal]

Cp = specific heat = 0.47 [cal/g*°C]

T_final = final temperature = 32 [°C]

T_initial = initial temperature = 27 [°C]

m = mass of the substance = 450 [g]

Now replacing:

Q=450*0.47*(32-27)\\Q=1057.5[cal]

5 0
2 years ago
Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f
musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

E_1_x = 0V/m\\E_1_y=100V/m

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m

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∑y: E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

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or 27.23° to the left of E1.

8 0
2 years ago
How dose an atom change if all of its electrons are removed
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Answer:

The atom becomes a positively charged ion.

Explanation:

  1. The building blocks of an atom are protons, neutrons, and electrons.
  2. The protons and neutrons present in the core of the atom are called nucleus.
  3. The electrons are scattered in an ordered way around the nucleus.
  4. The protons are positively charged and the electrons are negatively charged particles. The neutrons do not possess any charges.
  5. Binding energy is supplied to the atom to remove an electron.
  6. It is possible to remove the electrons of the lighter elements.
  7. When an electron is removed from the hydrogen atom. It becomes positively charged ion or simply proton.
  8. When all of the electrons are removed from the helium atom, it becomes a positively charged α particle.
  9. It is practically very difficult to remove all of the electrons from the heavier elements.
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