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jolli1 [7]
3 years ago
7

A cart travels 4.00 meters east and then4.00 meters north. determine the magnitude ofthe cart’s resultant displacement.

Physics
1 answer:
gogolik [260]3 years ago
7 0
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Square root of (4^2 + 4^2) = 4*squareRoot(2)
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A small 16 kilogram canoe is floating downriver at a speed of 4 m/s. What is the canoe's kinetic energy?
seraphim [82]

Kinetic Energy,K.E=1/2MV²

mass,m=16kg

velocity,v=4m/s

K.E=1/2×16×4²

=128kgm²/s²

=128 Joules

5 0
3 years ago
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8. Which explains how a plane mirror works?
lesya692 [45]

Answer:

light rays reflect off an object ,strike the mirror ,and are reflected into your eyes

6 0
2 years ago
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How
qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0
2 years ago
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A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret
emmasim [6.3K]

f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car  = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

f' = f (V + v)/(V - v)

inserting the values in the above equation

1250 = 1200 (340 + v)/(340 - v)

v = 6.9 m/s

4 0
3 years ago
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