All categories

2 years ago

How do I Change the following to decimal numbers. a. 5.93 x 10-5

Physics

1 answer:

PSYCHO15rus [73]2 years ago

5 0

Explanation:

To convert from scientific notation to decimal, move the decimal point 5 places to the left.

5.93×10⁻⁵ = 0.0000593

You might be interested in

A horse began running due east and covered 25 km in 4.0 hr. What is the average velocity of the horse?

Darina [25.2K]

3 0

3 years ago

Read 2 more answers

What is the magnitude of the velocity of a 25 kg mass that is moving with a momentum of 100 kg*m/s?

Gekata [30.6K]

Answer:

v= 4 m/s

Explanation:

Momenutm is, by definition, the product of mass and velocity.

$p = mv$

Let's replace what we know and solve for whatever's left

$100 kg\cdot m/s = 25kg \cdot v \rightarrow v= 4 m/s$

7 0

2 years ago

A stone is thrown at an angle of 30 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A

natta225 [31]

v = initial velocity of launch of the stone = 12 m/s

θ = angle of the velocity from the horizontal = 30

Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.

v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s

a = acceleration of the stone = - 9.8 m/s²

t = time of travel = 4.8 s

Y = vertical displacement of stone = vertical height of the cliff = ?

using the kinematics equation

Y = v₀ t + (0.5) a t²

inserting the values

Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²

Y = - 84.1 m

hence the height of the cliff comes out to be 84.1 m

5 0

2 years ago

A 5-g lead bullet traveling in 20°C air at 300 m/s strikes a flat steel plate and stops.

densk [106]

To solve this problem it is necessary to apply the concepts related to the Kinetic Energy and the Energy Produced by the heat loss. In mathematical terms kinetic energy can be described as:

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

Replacing we have that the Total Kinetic Energy is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (5*10^{-3})(300)^2$

$KE = 225J$

On the other hand the required Energy to heat up t melting point is

$Q_1 = mC_p \Delta T$

$Q_2 = L_f m$

Where,

m = Mass

$C_p =$ Specific Heat

$\Delta T =$ Change at temperature

$L_f =$ Latent heat of fussion

Heat required to heat up to melting point,

$Q = Q_1+Q_2$

$Q = mC_p \Delta T+L_f m$

$Q = 5*0.128*(327-20) + 5*24.7$

$Q = 310J$

The energy required to melt is larger than the kinetic energy. Therefore the heat of fusion of lead would be 327 ° C: The melting point of lead.

4 0

3 years ago

The weight of a rock is 8.5 pounds. Does this statement include

Kaylis [27]

Answer:

quantitative

Explanation:

4 0

2 years ago