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4 years ago

In physics what does 7.56 × 5.746 equal ?

Physics

1 answer:

lapo4ka [179]4 years ago

7 0

Answer:

43.43

Explanation:

5.746 x 7.56 = 43.43976

As the least number of desimal is two so our awnser should contain two digits after the decimal point.

Ans: 43.43.

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An athlete wants to refill an old punching bag used for training, as shown in the picture. When refilled, the bag will rip if it

In-s [12.5K]

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states action and reaction are equal and opposite. That is the force applied to an object is equal in magnitude to force experienced by the object but in opposite direction.

From the given question, when the bag exert a certain on the athlete, the athlete also exerts similar force to the bag but in opposite direction.

Thus, the complete sentence is as follows;

The maximum force that the athlete exerts on the bag is NEGATIVE 1,500 N and in the OPPOSITE DIRECTION of the force that the bag exerts on the athlete.

Learn more about Newton's third law of motion here: brainly.com/question/25998091

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2 years ago

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

So here <u>friction force on the given block will be same as its component on weight which is 218.55 N</u>

5 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

EleoNora [17]

Answer:

63.8 V

Explanation:

We are given that

$A_1=1.6 cm^2=1.6\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

$A_2=1.2 cm^2=1.2\times 10^{-4} m^2$

$A_3=4.4 cm^2=4.4\times 10^{-4} m^2$

$A_4=7 cm^2=7\times 10^{-4} m^2$

Potential difference,V=140 V

We know that

$R=\frac{\rho l}{A}$

According to question

$l_1=l_2=l_3=l_4=l$

In series

$R=R_1+R_2+R_3+R_4$

$R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})$

$R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})$

$R=\rho l(18284.6)$

$I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}$

Potential across 1.2 square cm= $V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V$

Hence, the voltage across the 1.2 square cm wire=63.8 V

3 0

3 years ago

In what sense did the flappers dressand dance represent a display of freedom

pishuonlain [190]

Answer:

Flappers are symbols of the Thundering Twenties, the social, political disturbance and expanded transoceanic social trade that followed the finish of World War I. Flappers were viewed as reckless for wearing inordinate cosmetics, drinking liquor, smoking cigarettes in broad daylight, driving vehicles and treating sex nonchalantly. They guaranteed that the flappers' dresses were 'close to exposure', 'nervy', 'foolish', and unintelligent.

6 0

3 years ago

HELP PLEASE NOWWWW !!!! ( 1 ) Jill and Scott both road their bikes for 30 minutes. Jill traveled 5 kilometers and Scott trave

katovenus [111]

Jill and Scott both moves for 30 minutes

now if Jill cover 5 km distance and Scott cover 10 km distance

now we know that the formula of speed is given as

$speed = \frac{distance}{time}$