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3 years ago

Does anyone get this? lol

Physics

1 answer:

nlexa [21]3 years ago

7 0

Answer:

(a). 640.000 --- 6 significant figures.

(b). 2.0 --- 2 significant figures.

(c). 0.520200 --- 6 significant figures.

(d). 10.05 --- 4 significant figures.

(e). 10,000.0 --- 6 significant figures.

(f). 30.7 --- 3 significant figures.

(g). 0.00000065 --- 8 significant figures.

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Experiment: Limestone

White raven [17]

When these bonds are destroyed, a reaction occurs. ... Vinegar reacting with limestone breaks the bonds of calcium carbonate and acetic acid.

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3 years ago

Which of the following statements are true? Select all that apply.

beks73 [17]

Answer:

The following options are true based on the properties of electric field;

a) Electric field lines near positive point charges radiate outward.

b) The electric force acting on a point charge is proportional to the magnitude of the point charge.

d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.

Explanation:

From option b) From coulomb's law F = Kq1q2r/r2

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3 years ago

Number 18 plz help physics question

Anvisha [2.4K]

Answer:

Explanation:

a) R = V/I = 0.7/0.022 = 32Ω

b) R = V/I = 0.6/0.005 = 120Ω

c) No...Current change is not linear with voltage change.

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2 years ago

A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th

Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

$Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

now we have

$W_{gravity} = -mg(Lsin\theta)$

$W_{friction} = -\mu mgcos(\theta) L$

so from above equation

$Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

so from above equation work done by external force is given as

$Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2$

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3 years ago

A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body

JulijaS [17]

We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
$S=30 m + 45m+65m=140 m$

while the total time of the motion is
$t=5 s+7s+5 s=17 s$

So, the average speed of the body is:
$v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s$

8 0

3 years ago