All categories

3 years ago

Chocolate chip cookies or chocolate fudge cookies?

Physics

1 answer:

Cerrena [4.2K]3 years ago

6 0

Chocolate chip cookies

You might be interested in

A wave travels at 18 m/s and has a wavelength of 3 meters. What is the frequency of the wave?

Anna35 [415]

Answer:

0.1667Hz

Explanation:

by explanation frequency is given by 1/time

after carrying out the calculation of time we can easily get frequency.

time=speed/distance

=18/3

=6s

but frequency=1/time

hence=1/6

=0.1667hertz

7 0

2 years ago

IMPORTANT ANSWER ALL 3 PLEASE!

frozen [14]

Answer:

4. Liters

5. Celsius

6. Grams

8 0

3 years ago

A stone is thrown vertically upward from the roof of the building. Does the position of the stone depend on the location chosen

mel-nik [20]

Yes it depend course you throw the stone at the roof it can't came course it in the roof

7 0

4 years ago

How do you find acceleration due to gravity with time and height given?

Feliz [49]

Here, height is given which will be the distance for a freely falling object.

The velocity will be

$v=\text{ }\frac{h}{t}$

and the acceleration will be

$a=\frac{v}{t}$

In this way, the formula works.

3 0

1 year ago

A certain freely falling object, released from rest, requires 1.95 s to travel the last 23.5 m before it hits the ground. (a) Fi

Ratling [72]

Answer:

(a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

Explanation:

Given that,

Time = 1.95 s

Distance = 23.5 m

(a). We need to calculate the velocity

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$-23.5=u\times1.95+\dfrac{1}{2}\times(-9.8)\times(1.95)^2$

$u=\dfrac{-23.5+4.9\times(1.95)^2}{1.95}$

$u=-2.496\ m/s$

(b). We need to calculate the total distance the object travels during the fall

Using equation of motion

$v = u+gt$

Put the value in the equation

$-2.496=0-9.8\times t$

$t =\dfrac{2.496}{9.8}$

$t=0.254\ sec$

The total time is

$t'=t+1.95$

$t'=0.254+1.95$

$t'=2.204\ sec$

We need to calculate the distance

Using equation of motion

$s = ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times9.8\times(2.204)^2$

$s=23.80\ m$

Hence, (a). The velocity of the object is -2.496 m/s.

(b). The total distance of the object travels during the fall is 23.80 m.

4 0

3 years ago