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Nataly_w [17]
3 years ago
13

you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic

ient of sliding friction is 0.18. How much work have you done when the box has risen 1meter vertically
Physics
1 answer:
Tomtit [17]3 years ago
5 0

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

W=Fd

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

d=\frac{h}{sin \theta} is the distance covered, where

h = 1 m is the vertical rise

\theta=30^{\circ} is the slope of the plane

Substituting and solving, we find

W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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A ball is thrown vertically upward with a speed of 19.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its
Arte-miy333 [17]

Answer with Explanation:

We are given that

Initial speed=u=19 m/s

a.g=9.8m/s^2

Final velocity of ball=v=0

v=u-gt

g is negative because the ball is going against to gravity.

0=19-9.8t

9.8t=19

t=\frac{19}{9.8}=1.94 s

s=ut-\frac{1}{2}gt^2

Using the formula

s=19(1.94)-\frac{1}{2}(9.8)(1.94)^2

s=18.4 m

a.The ball rise upto height 18.4 m

b.It take 1.94 s to reach its highest point.

c.Initial velocity=0,s=18.4 m

s=ut+\frac{1}{2}gt^2

18.4=0(t)+\frac{1}{2}(9.8)t^2

18.4=4.9t^2

t^2=\frac{18.4}{4.9}

t=\sqrt{\frac{18.4}{4.9}}

t=1.94 s

v=u+gt

Using the formula

v=0+9.8(1.94)=19 m/s

7 0
3 years ago
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A room has a heater on one side and
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Answer:

the warm air would be up and cool would be down

Explanation:

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Tim observes that a force pulls him down when he jumps up in the air. Based on his observation, which conclusion is correct?
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Gravitational force is a non-contact force.
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Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

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3 years ago
Which of the following elements is in the same period as phosphorus?
Arte-miy333 [17]
The answer is B. magnesium I am pretty sure
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3 years ago
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