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2 years ago

10

When the pressure at the bottom of a submerged object is greater than the pressure at the top of submerged object, buoyant force

is produced.
A. True
B. False

2 answers:

Ivanshal [37]2 years ago

8 0

Answer:

true!

Explanation:

ELEN [110]2 years ago

4 0

The answer would most like be true!

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topjm [15]

When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !

8 0

2 years ago

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Is there supposed to be a slit at the top of your adams apple?

Allushta [10]

No, there isn't. Please consult your doctor if this is the case with yours or someone you know.

7 0

2 years ago

A bicyclist bikes the 90 mi to a city averaging a certain speed. The return trip is made at a speed that is 1 mph slower. Total

Lynna [10]

Answer:

his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

Explanation:

Let say the speed of the bicycle while he moves towards the city is "v"

now the speed of the round trip must be smaller by 1 mph

so its speed for round trip will be

$v_2 = v - 1$

now we know that total time of the motion is 19 hr

so we will have

$t_1 = \frac{90}{v}$

$t_2 = \frac{90}{v - 1}$

so we will have

$t_1 + t_2 = 19 hr$

$\frac{90}{v} + \frac{90}{v-1} = 19$

$90(2v - 1) = 19(v^2 - v)$

$19 v^2 - 199 v + 90 = 0$

by solving above equation we have

$v = 10 mph$

so his speeds while going to city is 10 mph and while his round trip the speed will be 9 mph

5 0

2 years ago

A 500kg car skids to a stop at a traffic light, leaving behind a 18.25m skid mark as it comes to a rest. Assuming that the car i

Nastasia [14]

Answer:

Coefficient of friction will be 0.587

Explanation:

We have given mass of the car m = 500 kg

Distance s = 18.25 m

Initial velocity of the car u = 14.5 m/sec

As the car finally stops so final velocity v = 0 m/sec

From second equation of motion

$v^2=u^2+2as$

$0^2=14.5^2+2\times a\times 18.25$

$a=-5.76m/sec^2$

We know that acceleration is given by

$a=\mu g$

$5.76=\mu\times 9.81$

$\mu =0.587$

So coefficient of friction will be 0.587

6 0

3 years ago

How much heat energy is required to raise the temperature of 0.368kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of cop

Katarina [22]

The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.

Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal

3 0

2 years ago

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