Help please asap i give 30 pts

During a science fair, a group of students came up with the following question: “Is color an inherent property in objects or is

liraira [26]

I think the correct answer from the choices listed above is option B. The question that the group of students came up with would be an extremely difficult question to answer scientifically because how people see or experience color is subjective and people might see it differently.

3 0

3 years ago

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a football is thrown upward at a 31° angle to the horizontal.the acceleration of gravity is 9.8m/s^2. To throw the ball a distan

mr Goodwill [35]

The ball's vertical position $y$ in the air at time $t$ is

$y=v_0\sin31^\circ\,t-\dfrac g2t^2$

The ball is at its original height when $y=0$ , which happens at

$v_0\sin31^\circ\,t-\dfrac g2t^2=\dfrac t2\left(2v_0\sin31^\circ-gt)=0$

$\implies t=0\text{ and }t=\dfrac{2v_0\sin31^\circ}g$

Meanwhile, the ball's horizontal position $x$ at time $t$ is

$x=v_0\cos31^circ\,t$

So when the ball reaches its original height a second time, the ball will have traveled a horizontal distance of

$x=\dfrac{2{v_0}^2\sin31^\circ\cos31^\circ}g=\dfrac{{v_0}^2\sin(2\cdot31^\circ)}g$

(which you might recognize as the formula for the range of a projectile)

To reach a distance of $x=77\,\rm m$ , the initial speed $v_0$ would be

$77\,\mathrm m=\dfrac{{v_0}^2\sin62^\circ}{9.8\,\frac{\rm m}{\mathrm s^2}}\implies v_0=29\dfrac{\rm m}{\rm s}$

7 0

3 years ago

A 140 g baseball is moving horizontally to the right at 35 mis when it is hit by the bat. the ball fl ies off to the le ft at 55

Anit [1.1K]

Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

When it hits the bat, it flies left with a velocity of 55 m/s at an angle of 25°. To reiterate:

mass = 0.140kg, Initial: 35m/s, Final: 55m/s at an angle of 25°

For this problem, we have to use the impulse equation, but before that lets break the velocity into components (It will be apparent towards the end):

The initial velocity is moving only in the horizontal direction, so:

$v_{0x} = 35 m/s$

The final velocity has an x and a y component:

$v_{fx} = 55cos(25) = 49.84692829m/s\\v_{fy} = 55sin(25) = 23.2440044m/s$

Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

$J = dp= m*dv$

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

$dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s$

Finally, substitute into the equation

$J = m*dv\\J = 0.140kg * 87.97320604m/s\\J = 12.31624885kg*m/s\\J = 12.32 kg*m/s$

J = 12.32kg*m/s

8 0

3 years ago

A hill that has a 36.1% grade is one that rises 36.1 m vertically for every 100.0 ml of distance in the horizontal direction. At

GaryK [48]

Answer:

angle is 19.85°

Explanation:

given data

grade = 36.1 %

vertically distance y = 36.1 m

horizontal distance x = 100 m

to find out

angle

solution

we know that according to question here

it make a triangle shape that contain 100 m base and height 36.1 m

and make angle with base is θ

so by triangle we know here

tanθ = y/x ..........1

so we now put x and y and find θ

tanθ = 36.1 / 100

tanθ = 0.361

θ = 19.85°

so angle is 19.85°

7 0

3 years ago

Air in a 120 km/h wind strikes head on the face of abuilding 45m wide by 75m high and is brought to rest. if air has a mass of 1

AlekseyPX

From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
 v = 120/3.6 = 33.33 m/s
 dm/dt = v*area*density
 dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = 146235.375 kg/s
 F = v^2*area*density
 F = (33.33)^2*((45)*(75))*(1.3) = 4874025 N
 This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.

4 0

4 years ago

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