PLEASEE HELP I !!!!!!!!!!!!!!!!!!!!!!!!!

Select the correct answer. A ball is thrown straight down from the top of a building at a velocity of 16 ft/s. The building is 4

BartSMP [9]

Answer:

The ball takes 5s to reach the ground

Explanation:

in order to solve this problem we use the kinematics equation with gravity as acceleration:

$h=v_0t+1/2*gt^2$

we replace the values

$0=1/2*32t^2+16t-480$

We solve this quadratic equation:

t=5s

t=-6s (this solution has not physical sense)

8 0

3 years ago

First step in using a balance beam scale?

prohojiy [21]

To find out the weight of the object, you'll need to slide the weight poises until the pointer is at zero again. Start with the two heavier weight poises and then use the lightest one to do the fine tuning. Our triple beam balance and the Ohaus triple beam balance are accurate to 0.1 grams.

8 0

4 years ago

Is there any change in the pressure of container filled with water when the volumed is increased

marshall27 [118]

Not really the volume of a container is simply length X width X depth so just how big the container unless the water is pressurized by some sort of weight or if the containers air pressure is lowered

7 0

3 years ago

In a double-slit experiment, the third-order maximum for light of wavelength 490 nm is located 15 mm from the central bright spo

jeka94

Answer:

The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

Explanation:

Given that,

Wavelength = 490 nm

Distance y= 15 mm

Length L=1.6 m

New wavelength = 670

We need to calculate the distance from the central bright spot

Using formula of distance

$y= \dfrac{m\lambda L}{d}$

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{3\times490\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=156.8\times10^{-6}\ m$

$d=156.8\times10^{-3}\ mm$

We need to calculate the distance from the central bright spot for new wavelength

Using formula of distance

$d=\dfrac{m\lambda L}{y}$

Put the value into the formula

$d=\dfrac{2\times670\times10^{-9}\times1.6}{15\times10^{-3}}$

$d=142.9\times10^{-6} m$

$d=142.9\times10^{-3}\ mm$

Hence, The distance from the central bright spot are $156.8\times10^{-3}\ mm$ and $142.9\times10^{-3}\ mm$ .

8 0

3 years ago

A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler

Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

$W=K_f - K_i$

Since the cart was initially at rest, $K_i = 0$ , so

$W=K_f = \frac{1}{2}mv^2$ (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, $F_g = 99.5 N$ :

$m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg$

So solving eq.(1) for v, we find the final speed of the cart:

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s$

2) $2.51\cdot 10^7 J$

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

$F=4.28 \cdot 10^5 N$

$d=586 m$

So the work done is

$W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J$

3) $2.51\cdot 10^7 J$

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

$W=\Delta K = K_f - K_i$

where

W is the work done

$\Delta K$ is the change in kinetic energy

Therefore, the change in kinetic energy is

$\Delta K = W = 2.51\cdot 10^7 J$

4) 37.2 m/s

According to the work-energy theorem,

$W=\Delta K = K_f - K_i$

where

$K_f$ is the final kinetic energy of the train

$K_i = 0$ is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

$W=K_f = \frac{1}{2}mv^2$

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

$v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s$

7 0

4 years ago