All categories

3 years ago

6^5/2^3= what is the answer please

Physics

2 answers:

enot [183]3 years ago

7 0

Answer:

972

Explanation:

$\tt \cfrac{6^5}{2^3}= \cfrac{(2\cdot3)^5}{2^3}=\cfrac{2^5\cdot3^5}{2^3}=2^2\cdot3^5=4\cdot243=972$

kogti [31]3 years ago

6 0

Answer:

972

Explanation:

6^5/2^3

=7776/8

=972

You might be interested in

True or False? Mechanical waves do NOT need a medium to travel through.

stellarik [79]

Answer:

False.

Explanation:

Mechanical waves require a medium in order to transport their energy from one location to another.

6 0

3 years ago

Read 2 more answers

A 1.0 kg block is placed against an ideal spring with spring constant 800 N/m and initially compressed 0.20 m. The spring and bl

densk [106]

Answer:

the release will be at 3.266 m distance

Explanation:

mass = 1 Kg

spring constant (k) = 800 N/m

initial compression = 0.20 m

θ = 30⁰

$U= \dfrac{1}{2}kx^2\\U= \dfrac{1}{2}800\times 0.2^2\\U= 16 J$

$U=mgh\\h=\dfrac{U}{mg}\\h=\dfrac{16}{1 \times 9.8}\\h = 1.633m$

$d=\dfrac{h}{sin \theta}\\d=\dfrac{1.633}{sin 30}\\\\d= 3.266m$

hence the release will be at 3.266 m distance.

5 0

3 years ago

When your body is warmed by an electric blanket during the winter, this process is said to be

lana66690 [7]

The correct answer would be A. endothermic because endothermic is-<span>(of a reaction or process) accompanied by or requiring the absorption of heat. Hope this helps!:) If you need any more just tag me.

</span>

8 0

3 years ago

Read 2 more answers

Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load

rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

$\sigma=E*\epsilon$ (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$

$\sigma=22.44N / 1290 mm^2$

$\sigma=0.0174 N/mm^2$

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

$=\sigma=E*\epsilon$

$\epsilon=\sigma/E$

$\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2}$

$\epsilon=8.53*10^-{5}$

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

$\delta L=3048 mm * 8.53*10^{-5}$

$\delta L= 0.25 mm$

4 0

3 years ago

A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j

nika2105 [10]

To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,

$H = \frac{v_0^2sin^2\theta}{2g}$