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3 years ago

A student walks 160 m in 150. The student stops for 30and then walka 210 m farther in 140 What is the

Physics

1 answer:

mihalych1998 [28]3 years ago

7 0

Answer:

Average speed = 1.3 m/s

Explanation:

Given that a student walks 160 m in 150 s. The student stops for 30 s and then walka 210 m farther in 140 s.

Neglect the time the student stops.

The total distance covered will be

Distance = 160 + 210

Distance = 370 m

The total time of the whole distance journey will be

Time = 150 + 140 = 290 s

The formula for speed is

Speed = distance/time

Speed = 370/290

Speed = 1.276 m/s

Therefore, the average speed is 1.3 m/s approximately

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A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surfac

zhuklara [117]

Answer:

W=76.55 miles.metric tons

Explanation:

Given that

Weight on the earth = 12 tons

So weight on the moon =12/6 = 2 tons

( because at moon g will become g/6)

As we know that

$F=\dfrac{K}{x^2}$

Here x= 1100 miles

F 2 tons

$2=\dfrac{K}{1100^2}$

$K=2.4\times 10^6$

We know that

Work = F. dx

$W=\int_{x_1}^{x_2}F.dx$

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$W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}$

$W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]$

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6 0

3 years ago

The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries

tester [92]

Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance = 2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W

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3 years ago

A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and

mel-nik [20]

Answer:

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Explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion

$\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\$