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wlad13 [49]
3 years ago
7

Calculate the speed of your cat as it runs towards its food bowl 14.7m away in 4.5 s. Give answer in mph.

Physics
1 answer:
Luden [163]3 years ago
4 0

Answer:

3.26mph

Explanation:

To calculate speed, use the formula distance/time. In this case, just divide 14.7  by 4.5.

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A lamp consumes 1000J of ekectrical energy in 10s. Calculate its power.​
Mice21 [21]
You do 1000 divide it by 10 which equals 100 W
4 0
3 years ago
Use the graph of velocity versus time for an object to answer the question.
Pavel [41]

T<u>he statement which fairly compares segment 2 and segment 3 </u>is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Since segment 2 starts at t = 60 s and ends at t = 150 s, the time interval is Δt = 150 - 60 = 90 s.

Also, segment 3 starts at t = 150 s and ends at t = 240 s, the time interval is Δt = 240 - 150 = 90 s.

So, their time periods are the equal.

We notice that segment 2 is less steep than segment 3 this implies that the acceleration in each segment is different, since the acceleration is the slope of the graph.

Since force is determined by acceleration, this implies that the force on segment 2 is different form the force acting in segment 3.

So, we have equal time periods but different forces.

So, <u>the statement which fairly compares segment 2 and segment 3 </u>is These represent equal periods of time, but the force during segment 2 is different than the force during segment 3.

Learn more about velocity-time graph here:

brainly.com/question/24788847

8 0
3 years ago
Grade 12 math: vectors as forces?
sashaice [31]
Larger tug force = 2*cos20 forward + 2*sin20 to the right (starboard) 
<span>Smaller tug force = 1*cos15 forward + 1*sin15 to the left (port) </span>

<span>Resultant force forward = 2*cos20 + 1*cos15 = 2.8453 (4dp) </span>
<span>Resultant force to right = 2*sin20 - 1*sin15 = 0.4252 (4dp) </span>

<span>Angle of resultant force = arctan(0.4252 / 2.8453) = arctan(0.1494) = 8.5 degrees</span>
4 0
3 years ago
Lake Erie contains roughly 4.00 ✕ 1011 m3 of water. (a) How much energy is required to raise the temperature of that volume of w
denis-greek [22]

Answer:

Part a)

Q = 6.36 \times 10^{18} J

Part b)

t = 144.11 years

Explanation:

Part a)

Energy required to raise the temperature of water from 17.8 degree C to 21.6 degree C is given by formula

Q = ms\Delta T

here we know that

Here the volume of the water is given as

V = 4.00 \times 10^{11} m^3

now the mass of water is

m = density \times volume

m = 4.00 \times 10^{14} kg

now the heat required is

Q = (4 \times 10^{14})(4186)(21.6 - 17.8)

Q = 6.36 \times 10^{18} J

Part b)

As we know that power is supplied at

P = 1400 MW

so here we know

P = \frac{Q}{t}

so here we have

t = \frac{Q}{P}

t = \frac{6.36 \times 10^{18}}{1400 \times 10^6}

t = 4.54 \times 10^9 s

t = 144.11 years

7 0
3 years ago
The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo
Advocard [28]

Answer:

a) the angular speed of the Earth's rotation is <em>7.272 × 10⁻⁵ rad/s.</em>

<em></em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

<em></em>

c) Earth's speed at a point on the equator is <em>463.23 m/s.</em>

<em></em>

d)  the speed of a point on the earth’s surface halfway between the equator and the pole is <em>231.61 m/s.</em>

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

<em>ω = 7.272 × 10⁻⁵ rad/s</em>

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the <em>angular velocity is positive.</em>

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 463.23 m/s</em>

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

<em>v = 231.61 m/s</em>

6 0
3 years ago
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