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3 years ago

10

assume that in an unpolished floor a lady us pushing a 40 kg box to the left with an applied force of 192 N. find the sum of for

ce

1 answer:

Vesnalui [34]3 years ago

5 0

The sum of force is 7,680...............

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As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near the center o

MrRa [10]

Answer:

18.6012339739 A

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

L = Length of wire = 55 cm

N = Number of turns = 4000

I = Current

Magnetic field is given by

$B=\dfrac{\mu_0NI}{L}\\\Rightarrow I=\dfrac{BL}{\mu_0N}\\\Rightarrow I=\dfrac{0.17\times 0.55}{4\pi \times 10^{-7}\times 4000}\\\Rightarrow I=18.6012339739\ A$

The current necessary to produce this field is 18.6012339739 A

7 0

3 years ago

A square plate of edge length 9.0 cm and negligible thickness has a total charge of 6.90 10-6 C. Estimate the magnitude E of the

SVETLANKA909090 [29]

Answer:

E= 4.35*10^6 N/C

Explanation:

Let's find the area charge density of the plate

α= 6.9*10^-6/9*10^-2 = 7.7*10^-5C/m2

Now we can calculate the electric field just of the plate

E =α/2e =7.7*10^-5/2*8.85*10^-12 = 4.35*10^6 N/C

7 0

3 years ago

When was the most gravitational potential energy stored between the model and Earth? Assume that the model's mass did not change

Lyrx [107]

Answer:

As the ball falls from C to E, potential energy is converted to kinetic energy. The velocity of the ball increases as it falls, which means that the ball attains its greatest velocity, and thus its greatest kinetic energy

Explanation:

8 0

3 years ago

An ideal gas is in a sealed rigid container. the average kinetic energy of the gas molecules depends most on

finlep [7]

It depends most on the temperature of the gas.

8 0

3 years ago

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

3 years ago