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3 years ago

How much is required to accelerate a 5kg mass at 20 m/s2

Physics

1 answer:

zhannawk [14.2K]3 years ago

5 0

use F = ma

F : force m : mass a : acceleration

f = 5kg * 20 m/s2 = 100 N

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Severus Snape knows that density of his powder is 3.00 g/cm3 . He also knows he needs 3.00 cm3 of this powder. What mass in gram

harina [27]

Answer:

He requires 1 gram of mass.

Explanation:

The density is defined as:

$\rho = \frac{m}{V}$ (1)

Where m is the mass and V is the volume.

Then, m can be isolated from equation 1 in order to determine the mass.

$m = \rho \cdot V$ (2)

$m = (3.00g/cm^{3})(3.00cm^{3})$

$m = 1g$

Hence, he requires 1 gram of mass.

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3 years ago

Two identical twins hold on to a rope, one at each end, on a smooth, frictionless ice surface. They skate in a circle about the

777dan777 [17]

Answer:

Part a)

$L = 2683.2 kg m^2/s$

Part b)

$v' = 8.60 m/s$

Part c)

$W = 4326.7 J$

Explanation:

Part a)

As we know that there is no external torque on the system of two twins

so here we will use

$L = mv r + mvr$

$L = 2(78 \times 4.30 \times 4)$

$L = 2683.2 kg m^2/s$

Part b)

Since angular momentum is conserved here as there is no external torque

so we will have

$2(m v r) = 2( m v' \frac{r}{2})$

$v' = 2v$

$v' = 8.60 m/s$

Part c)

Work done by both of them = change in kinetic energy

so we have

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

$W = m(v'^2 - v^2)$

$W = 78(8.60^2 - 4.3^2)$

$W = 4326.7 J$

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3 years ago

Write a linear equation that expresses the relationship between the temperature in degrees Celsius (C) and degrees Fahrenheit (F

UkoKoshka [18]

Answer:

$\displaystyle F=\frac{9}{5}C+32$

$\displaystyle C=\frac{5}{9}(F-32)$

Explanation:

Temperature Units Conversion

The conversion formula between Celsius and Fahrenheit temperature scales is well-known. But we'll use the provided data to derive the formula. Let's model the relationship between Fahrenheit (F) and Celsius (C) as a linear function like

$F=mC+b$

Where m and b must be computed according to the pair of conditions given. The values for each temperature scale are (C,F)=(0,32) and (100,212). Replacing the first value

$32=m\times 0+b$