Students set up a controlled experiment (please help I am so tired and unable to function lol)

During the last shot of the game, the basketball goes from rest to 15 m/s and reaches the backboard in 0.41 s. What was the acce

Anna [14]

Answer: a= 37m

Explanation: V= 15 m/s (Velocity) t= 0.41s (time) formula: a= v/t

15 m/s / 0.41 (15 divided by 0.41) = 36.583m

There are 2 significant digits, 36, you look at the third digit, either round up or down in this case up to 36. a= 37m

5 0

3 years ago

Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi

Dimas [21]

Answer:

Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker

D2 = 5.05 m from the far speaker

The difference in distances from the speakers is

5.05 - 4.5 = .55 m (Let y be wavelength, lambda)

n y = 4.5

(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength

y = .55 m subtracting equations

f = v / y = 340 / .55 = 618 / sec should be the smallest frequency

8 0

2 years ago

Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i

aliya0001 [1]

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

8 0

2 years ago

Calculate the Force of Gravity acting on an object that has a mass of 1.3 kg. The object is on Earth so use 9.8m/s2 for the acce

sweet-ann [11.9K]

To calculate the gravitational force acting on an object given the mass and the acceleration due to gravity, use the following formula.

Fg = m • g
Fg = 1.3 kg • 9.8 m/s^2
Fg = 12.74 N or about 12.7 N.

The solution is C. 12.7 N.

7 0

2 years ago

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

8 0

3 years ago