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erik [133]
2 years ago
14

A pizza delivery person drives 3 miles North then 5 miles West and then 3 miles South to deliver a pizza. What is the person's d

isplacement?
A. 45 miles
B. 8 miles
C. 11 miles
D. 5 miles west
Physics
2 answers:
Anton [14]2 years ago
7 0

Answer: it’s 11

Explanation:

I just did the question

AleksAgata [21]2 years ago
4 0

Answer:11

Explanation:

Just did it

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WORD BANK:
Mkey [24]

Answer:

1. Vector, base

2. Vector, derived

3. Vector, ?

4. Scalar, derived

5. scalar, base

4 0
2 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
Igoryamba

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

-----------------------

Note: we don't use the mass at all

6 0
3 years ago
New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

  • Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

4 0
2 years ago
A 0.660 kg ball is dropped from rest from the top of a building and falls for 5.65 seconds. How tall was the building ?
kirill115 [55]

Answer:

The height of the building is approximately 156.58 m

Explanation:

The mass of the ball dropped from rest from the building top = 0.660 kg

The time in which the ball falls, t = 5.65 seconds

The height, h, of the building is given from the following equation of motion;

h = u·t + ¹/₂·g·t²

Where;

u = The initial velocity of the ball = 0 m/s

g = The acceleration due to gravity = 9.81 m/s²

Plugging in the values, we have;

h = 0 × 5.65 + ¹/₂ × 9.81 × 5.65² ≈ 156.58 m

The height of the building, h ≈ 156.58 m.

6 0
3 years ago
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