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sergiy2304 [10]

3 years ago

14

Please help on this one?

2 answers:

Tpy6a [65]3 years ago

8 0

C reactant. Both 2C0 and 02 are reactants and 2C02 is the product

nikdorinn [45]3 years ago

4 0

a is the right answer

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Arunning back practices for his upcoming football game by running drills. He runs forward 62.4 yards, and runs backward for

Mama L [17]

Answer: He has only move 0.2 yards

Explanation: When you subtract 18.3 from 18.5 you get 0.2 and that is how much he's moved

4 0

3 years ago

Which can be used as a lever ?

Svet_ta [14]

B long board because of the length

7 0

3 years ago

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav

Dmitrij [34]

Answer:

$t=1.5\times 10^{-4}\ s$

Explanation:

Given:

file size to be transmitted, $D=30\ Mb$

transmission rate of data, $\dot D=10\ Mb.s^{-1}$

propagation speed, $v=2\times 10^8\ m.s^{-1}$

distance of data transfer, $s=10000\ km=10^4\ m$

<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

$t'=\frac{s}{v}$

$t'=\frac{10^4}{2\times 10^8}$

$t'=5\times 10^{-5}\ s$

<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

$t=3\times t'$

$t=3\times 5\times 10^{-5}$

$t=1.5\times 10^{-4}\ s$

3 0

4 years ago

The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers.

AVprozaik [17]

Answer:

F = 1.63 x 10⁻⁹ N

Explanation:

Complete question is as follows:

The diagram below shows two bowling balls, A and B, each having a mass of 7.0 kg, placed 2.00 m apart between their centers. Find the magnitude of Gravitational Force?

Answer:

The gravitational force is given by Newton's Gravitational Law as follows:

F = Gm₁m₂/r²

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of each ball = 7 kg

r = distance between balls = 2 m

Therefore,

F = (6.67 x 10⁻¹¹ N.m²/kg²)(7 kg)(7 kg)/(2 m)²

<u>F = 1.63 x 10⁻⁹ N</u>

7 0

3 years ago

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

$K_t = \frac{1}{2} * m * v^2$

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

So

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

3 0

3 years ago