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2 years ago

A chemical reaction’s speed can be increased by ____.

Physics

2 answers:

fenix001 [56]2 years ago

6 0

A. Using a catalyst

Oxana [17]2 years ago

5 0

The answer is A) Using a catalyst

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What must the charge (sign and magnitude) of a 3.45 g particle be for it to remain stationary when placed in a downward-directed

Pani-rosa [81]

charge must be equal to 5.74 ×10⁻⁵

In the question it is said that the particle remains stationary which means the the net force on the particle is zero. So, the counterbalancing forces must be equal which means weight is equal to upward electric force.

→ Fnet =0

→ mg = qE

substituting the values we get :

0.00345 × 9.81 = q × 590

→ q = 5.74 ×10⁻⁵

Hence the charge must be equal to 5.74 ×10⁻⁵.

Learn more about charges here:

brainly.com/question/26092261

# SPJ4

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1 year ago

If i ask wat r the application of simple machine some of u will say-----> Simple machines that are widely used include the wh

TiliK225 [7]

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2 years ago

2 resistors of resistance 1000 ohm and 2000 ohm are joined in series with a 100V supply. A voltmeter of internal resistance 4000

Vadim26 [7]

<h2>The voltmeter reading will be 35.7 volt </h2>

Explanation:

The resistor 1000 ohm and 4000 ohm are connected in parallel .

Their combined resistance is supposed R₁

Thus $\frac{1}{R_1}$ = $\frac{1}{1000}$ + $\frac{1}{4000}$

or R₁ = 800 ohm

Therefore the total resistance in circuit is = 2000 + 800 = 2800 ohm

Because these are in series .

We can find current flowing through the circuit I = $\frac{V}{R}$ = $\frac{100}{2800}$ = $\frac{1}{28}$

here R is total resistance .

The potential difference across 1000 ohm = $\frac{1}{28}$ x 1000 = 35.7 volt

Thus voltmeter reading will be 35.7 volt

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2 years ago

tankabanditka [31]

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3 years ago

The speed of a tennis ball that is served is 73.14 m/s. During a serve, the ball typically starts from rest and is in contact wi

Serga [27]

Answer: Third option

F = 250w

Explanation:

The impulse can be written as the product of force for the time interval in which it is applied.

$I = F (t_2-t_1)$

You can also write impulse I as the change of the linear momentum of the ball

$I = mv_2 -mv_1$

So:

$F (t_2-t_1) = mv_2 -mv_1$

We want to find the force applied to the ball. We know that

$(t_2-t_1) = 30$ milliseconds = 0.03 seconds

The initial velocity $v_1$ is zero.

The final speed $v_2 = 73.14\ m / s$

$F * 0.03 = 73.14m$

$F * 0.03 = 73.14m\\\\F=\frac{73.14m}{0.03}\\\\F=2438m$

We must express the result of the force in terms of the weight of the ball.

We divide the expression between the acceleration of gravity

$g = 9.8\ m / s ^ 2$

$F=\frac{2438m*g}{g},\ \ m*g=w\\\\g=9.8\ m/s^2\\\\F=\frac{2438w}{9.8}\\\\F=249w$

The answer is the third option

3 0

3 years ago