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n200080 [17]
3 years ago
14

A change in position is calleda ) speed b) distance ​

Physics
2 answers:
Irina-Kira [14]3 years ago
6 0
A its a its a its a its a
amid [387]3 years ago
6 0

Answer:

<h3>(b) distance </h3>

as well as displacement

Explanation:

<u>Distance</u> is a scalar quantity that refers to "how much ground an object has covered" during its motion.

<u>Displacement</u><u> </u>is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
Brrunno [24]

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector X = 7.8*Cos(50) = 5.01 Km y y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km) so that Tan (\alpha ) = \frac{0.375}{5.01} = 4.28; \alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree) to get how far we use Pythagorean theorem so R^{2} = x^{2}+y^{2} so that R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)

6 0
3 years ago
Read 2 more answers
Projectile's horizontal range on level ground is R=v20sin2θ/g. At what launch angle or angles will the projectile land at half o
seraphim [82]

Answer:

\theta = 15^o \: or\: 75^o

Explanation:

As we know that the formula of range is given as

R = \frac{v^2sin2\theta}{g}

now we know that

maximum value of the range of the projectile is given as

R_{max} = \frac{v^2}{g}

now we need to find such angles for which the range is half the maximum value

so we will have

\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}

sin(2\theta) = \frac{1}{2}

2\theta = 30 or 150

\theta = 15^o \: or\: 75^o

7 0
3 years ago
A very smart 3-year-old child is given a wagon for her birthday. She refuses to use it. "After all," she says, "Newton's third l
natita [175]

Answer:

Explanation:

She's correct but doesn't mean the wagon cannot put into motion. The force that she applied on the wagon, according to Newton's 2nd law, would have generated an acceleration, which translates into motion. The reaction force the wagon applies on her due to Newton's 3rd law, would not hinder its own motion.

5 0
3 years ago
Suppose the half-life of an element is 10 years. How many half-lives will it take before only about 6% of the original sample re
sergiy2304 [10]
You don't need to worry about the 10 year bit with this question. Just grab a calculator and divide 100/2, then the answer to that (50) by 2 etc and keep dividing by 2 until you get down to 6.25.

The answer ends up being 4 half lives :)

If you don't understand what a half life is please let me know :)
4 0
4 years ago
Read 2 more answers
Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius
Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2

The centripetal acceleration for the second radius; 4.0 m

a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2

The centripetal acceleration for the third radius; 6.0 m

a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2

The centripetal acceleration for the fourth radius; 8.0 m

a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2

The centripetal acceleration for the fifth radius; 10.0 m

a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2

6 0
3 years ago
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