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3 years ago

What is the weight of a 63.7 kg person?

Physics

2 answers:

dybincka [34]3 years ago

7 0

Answer: 140.4 pounds

Explanation:

IRINA_888 [86]3 years ago

6 0

Answer:

A person that weighs 63.7 kg weighs 63.7 kg

Explanation:

Too complicated to explain.

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A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0

3 years ago

The rate of a reaction is the speed at which products form or reactants disappear.<br> T/F

Natali [406]

Answer: TRUE (:

Explanation:

3 0

3 years ago

A parallel-plate capacitor in air has circular plates of radius 3.0 cm separated by 1.1 mm. Charge is flowing onto the upper pla

Yuki888 [10]

Answer:

2 x 10^14 N/Cs

Explanation:

radius, r = 3 cm

Area , A = 3.14 x 3 x 3 = 28.26 cm^2 = 28.26 x 10^-4 m^2

d = 1.1 mm = 1.1 x 10^-3 m

i = 5 A

Let time be t and electric field strength is E.

Charge, q = i x t = 5 t

q = C V

q = C x E x d

$5 t = \frac{\varepsilon _{0}A}{d}\times E\times d$

$E/t = \frac{5}{\varepsilon _{0}A}$

$E/t = \frac{5}{8.854\times 10^{-12}\times 28.26\times 10^{-4}}$

E/t = 2 x 10^14 N/Cs

5 0

3 years ago

Two charges, one +Q and the other −Q, are held a distance d apart. Consider only points on the line passing through both charges

solong [7]

Answer:

a. d/2 mid-way between the charges.

b. d/2 mid-way between the charges.

Explanation:

(a) Find the location of all points, if any, where the electric potential is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric potential due to the +Q charge is V = kQ/x and that due to the -Q charge is V' = -kQ/(d - x) where x is the point of zero electric potential.

The potential is zero when V + V' = 0, and this can only be midway between the charges. This is shown below

So, kQ/x + [-kQ/(d - x)] = 0

kQ/x - kQ/(d - x) = 0

kQ/x = kQ/(d - x)

1/x = 1/(d - x)

(d - x) = x

d = x + x

d = 2x

x = d/2 which is mid-way between the charges.

(b) Find the location of all points, if any, where the electric field is zero.

Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric field due to the +Q charge is E = kQ/x² and that due to the -Q charge is E' = -kQ/(d - x)² where x is the point of zero electric field.

The electric field is zero when E + E' = 0 and this can only be midway between the charges. This is shown below.

So, kQ/x² + [-kQ/(d - x)²] = 0

kQ/x² - kQ/(d - x)² = 0

kQ/x² = kQ/(d - x)²

1/x² = 1/(d - x)²

(d - x)² = x²

d - x = ± x

d = x ± x

d = x - x or x + x

d = 0 or 2x

d = 0 or d = 2x

Since d ≠ 0, d = 2x ⇒ x = d/2 which is midway between the charges.

4 0

3 years ago

How images from space of earth and other objects are useful

elena55 [62]

To show stuff that we cant see in very well

3 0

3 years ago