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Pepsi [2]
3 years ago
7

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

the length of the athlete's jump?
Physics
1 answer:
Lemur [1.5K]3 years ago
5 0

Answer:

      R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

           R = v₀² sin  2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

           R = 9.14² sin (2  35) / 9.8

           R = 8.01 m

this is the jump length

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A block of mass 4 kg slides down an inclined plane inclined at an angle of 30o with the horizontal. Find the acceleration of the
Nataly [62]

Answer:

a) a = 4.9 m/s²

b) a = 1.5 m/s²

Explanation:

no friction

F = ma

gsinθ = ma

a = gsinθ

a = 9.8sin30

a = 4.9 m/s²

friction

gsinθ - μmgcosθ = ma

a = g(sinθ - μcosθ)

a = 9.8(sin30 - 0.4cos30)

a = 1.5051...

8 0
3 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
3 years ago
A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
Rus_ich [418]

Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\

As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
3 years ago
1. A student practicing for a track meet ran 263 m in 30 sec. What was her average speed?
faltersainse [42]

Answer: By 47

Explanation: subtract 310 and 263

6 0
2 years ago
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of
pentagon [3]

Answer:

7772.72N

Explanation:

When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page  (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.

Now which direction is the static friction, assume that it is pointing inward so

Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N

Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.

7 0
3 years ago
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