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3 years ago

1. What shape is the Pencil's body? a. Square c. Circle b. Triangle d. Rectangle

Physics

2 answers:

Juliette [100K]3 years ago

4 0

<h3>Answer: <em><u>C circle </u></em></h3><h3><em><u /></em></h3>

Explanation:

AURORKA [14]3 years ago

4 0

C it’s a circle!!!!!

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BMX bicycle rider takes off from a ramp at a point 1.2 m above the ground. The ramp is angled at 40∘ from the horizontal, and th

podryga [215]

Answer:

The ramp lands at an horizontal distance of 3.989 m

Explanation:

Range: Range is defined as the horizontal distance of a projectile from the point of projection to the point where the projectile hits the projection plane again. It is measure in meters (m)

R = U²sin2∅/g..................................... Equation 1

Where R = The horizontal distance from the end of the ramp, U = rider's speed, ∅ = the ramp's angle to the horizontal.

<em>Given: U = 6.3 m/s, ∅ = 40°, g = 9.8 m/s²</em>

<em>Substituting these values into equation 1</em>

R = 3.989 m

Thus the ramp lands at an horizontal distance of 3.989 m

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3 years ago

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Vilka [71]

Answer:

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Explanation:

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3 years ago

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What is the definition of physical weathering?

jeyben [28]

Answer rocks, soil and minerals being slowly broken down or broken apart by the Earth's environment such as pressure, temperature, water and ice

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4 years ago

A car traveling in a straight line has a velocity of +4.4 m/s. after an acceleration of 0.65 m/s2 , the car's velocity is +8.3 m

nydimaria [60]

Answer:

1.53seconds

Explanation:

Using first equation of motion :

V=U + at

Where final velocity (V) =+8.3m/s

Initial velocity (U) =+4.4m/s

Acceleration (a) = 0.65m/s^2

time(s)=?

V=U + at

+8.3^2 = +4.4 + 0.65 * t

Making t the subject of the formula :

Therefore, t= ( +8.3 - 4.4)/0.65 = 1.53seconds

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4 years ago

Tarzan, who weighs 849 N, swings from a cliff at the end of a 18.0 m vine that hangs from a high tree limb and initially makes a

loris [4]

Answer:

Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

$F = 384.2 N$

Part d)

$\theta = 333 degree$

Part e)

$a = 4.4 m/s^2$

Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

$T = 342.5 \hat i + 675\hat j$

Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

6 0

3 years ago