All categories

2 years ago

75% of what number is 27? EXPLAIN

Physics

2 answers:

zmey [24]2 years ago

8 0

75 percent (calculated percentage %) of what number equals 27? Answer: 36.

Alchen [17]2 years ago

6 0

36 is the answer. Hope this helps!!!

You might be interested in

How is carbonic acid responsible for the formation of some caves? ASAP PLEASE!!!!

butalik [34]

It desolves calcium carbonate causing plates to bend and faults to occur. Faults and plates moving cases the formation of caves.

-Hopes this helps :)

4 0

3 years ago

A 4.33 kg cat has 41.7 J of KE How fast is the cat moving?

balandron [24]

Answer:

The answer to your question is:

Explanation:

Data

mass = 4.33 kg

E = 41.7 J

v = ?

Formula

Ke = (1/2)mv²

Clear v from the equation

v = √2ke/m

Substitution

v = √2(41.7)/4.33

v = 19.26 m/s Result

7 0

3 years ago

A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f

Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

$T=\dfrac{2\pi}{\omega}$

${\omega}=\dfrac{2\pi }{0.2}\ rad/s$

${\omega}=31.41\ rad/s$

We know that

${\omega}^2=m\ K$

K=Spring constant

$K=\dfrac{\omega^2}{m}$

$K=\dfrac{31.41^2}{0.12}\ N/m$

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0

3 years ago

Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through

liraira [26]

Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

= 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

4 0

3 years ago

Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

nydimaria [60]

Answer:

Explanation:

The formula for this is

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

$7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}$ and moving things around to solve for r:

$r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }$ Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

8 0

3 years ago