The area of the bar over r = 2 is 0.234. what is the area of the bar over r = 4?

In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho

Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance R horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

$R=u_xt=(ucos \theta)t$

Therefore,

$t=\frac{R}{ucos\theta} .......(1)$

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

$y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2$

Substitute the value of t from equation (1).

$y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )$

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

$y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s$

The shot put was thrown with a speed 15.02 m/s.

7 0

2 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

7 0

2 years ago

Read 2 more answers

When Kevin pulls his cotton shirt off his body, the electrons get transferred from the (shirt or body) to the (shirt or body) .

Masja [62]

When Kevin pulls his cotton shirt off his body, the electrons get transferred from the shirt (in form of static charges i.e. electrons to the body. So, the shirt becomes positively charged and Kevin’s body becomes negatively charged.

As a result of charge transfer from the shirt to the body, we can hear a crackling sound. or if observed in dark, a sparkle can be seen.

6 0

3 years ago

Carter pushes a bag full of basketball jerseys across the gym floor. The he pushes with a constant force of 21 newtons. If he pu

Zolol [24]

Carter needs a power equal to 63 W to be able to push the bag full of Jersey. This is by using the formula: Power is equal to the product of Force applied and Displacement all over time traveled.

8 0

3 years ago

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19. When baking soda and vinegar are combined, the result is a salt called sodium acetate. This forms

Orlov [11]

It is a chemical change because the baking soda and vinegar are reacting to form a new product.

3 0

3 years ago