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Marina86 [1]
3 years ago
7

a bus travels 4 km due north and 3 km due west going from bus station a to bus station b. the magnitude of the bus displacement

from station a to station b is what km
Physics
1 answer:
ladessa [460]3 years ago
4 0

Answer:

5km

Explanation:

Magnitude of displacement is found by getting the resultant. Resultant is same as the bypotenuse hence

R=\sqrt {{x^{2}+y^{2}} where x is the displacement in west direction and y is displacement in North direction. Substituting x with 3km and y with 4 km then

R=\sqrt {{3^{2}+4^{2}}=5km

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Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of
Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0
3 years ago
A rock thrown with speed 8.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 19.0 m bef
frez [133]
Draw a diagram to illustrate the problem as shown below.

The vertical component of the launch velocity is
v = (8.5 m/s)*sin30° = 4.25 m/s
The horizontal component of the launch velocity is
8.5*cos30° = 7.361 m/s

Assume that aerodynamic resistance may be ignored.
Because the horizontal distance traveled is 19 m, the time of travel is
t = 19/7.361 = 2.581 s

The downward vertical travel is modeled by
h = (-4.25 m/s)*(2.581 s) + 0.5*(9.8 m/s²)*(2.581 s)²
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Answer: The height is 21.7 m (nearest tenth)

4 0
3 years ago
A pear hangs in a tree at a height of 1.8 m. The pear has a mass of 0.2 kg. The pear falls out of the tree and lands on the grou
TiliK225 [7]

a) PE=mgh=0.2*9.8*1.2=2.352 J

b) KE=PE=2.352 J

c) v=\sqrt{\frac{2KE}{m}}=4.85 m/s

6 0
3 years ago
Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart n
Elanso [62]

Answer:

λ = 6.602 x 10^(-7) m

Explanation:

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is given as ;

y = mλD/d

Where;

D is the distance of the screen from the slits = 6.2 m

d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m

The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.

Therefore, from the previous formula we can find the wavelength of the light:

y = mλD/d

So, λ = dy/mD

Thus,

λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)

λ = 6.602 x 10^(-7) m

8 0
3 years ago
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