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3 years ago

Why is there an upward force on objects in water?

2 answers:

6 0

From pressure, so if you had a pot and turned it upside down and put it in water then turned it on its side the air bubbles would come out but if you kept trying to push it down it would get harder just like an inflatable boat if u turned it upside down tried to pull it under water you couldnt because of the pressure hope this helps can u mark brainliest please

prisoha [69]3 years ago

3 0

This is due to water pressure

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It took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 km/h. How far did the train travel

Nadusha1986 [10]

Answer: 420 km

Explanation:

120 per hour, 3.5 hours

120 x 3.5 = 420

6 0

3 years ago

True or False an overtone occurs when two or more sound waves are produced at the same time.

Setler79 [48]

Answer:

True, overtone occurs when two or more sounds are produced at the same time.

5 0

3 years ago

Read 2 more answers

Children in a tree house lift a small dog in a basket 4.00 m up to their house. If it takes 187 J of work to do this, what is th

Vika [28.1K]

A=Fh
A - work
F - force
h - distance

F=mg
m - mass (god+basket)

so
A=mgh
187 = m*10*4
187=40m
m=187/40
m=4.675 kg
or 4kg and 675g

pretty small dog...

6 0

3 years ago

Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.8 s after the eleva

Delicious77 [7]

Answer:

v= 4.0 m/s

Explanation:

When standing on the bathroom scale within the moving elevator, there are two forces acting on Henry's mass: Normal force and gravity.
Gravity is always downward, and normal force is perpendicular to the surface on which the mass is located (the bathroom scale), in upward direction.
Normal force, can adopt any value needed to match the acceleration of the mass, according to Newton's 2nd Law.
Gravity (which we call weight near the Earth's surface) can be calculated as follows:

$F_{g} = m*g = 95 kg * 9.8 m/s2 = 930 N (1)$

According to Newton's 2nd Law, it must be met the following condition:

$F_{net} = F_{g} -F_{n} = m*a\\ F_{net} = 930 N - 830 N = 100 N = 95 Kg * a$

As the gravity is larger than normal force, this means that the acceleration is downward, so, we choose this direction as the positive.

Solving for a, we get:

$a =\frac{F_{net} }{m} =\frac{100 N}{95 kg} = 1.05 m/s2$

We can find the speed after the first 3.8 s (assuming a is constant), applying the definition of acceleration as the rate of change of velocity:

$v_{f} = a* t = 1.05 m/s * 3.8 m/s = 4.0 m/s$

Now, if during the next 3.8 s, normal force is 930 N (same as the weight), this means that both forces are equal each other, so net force is 0.
According to Newton's 2nd Law, if net force is 0, the object is either or at rest, or moving at a constant speed.
As the elevator was moving, the only choice is that it is moving at a constant speed, the same that it had when the scale was read for the first time, i.e., 4 m/s downward.

3 0

3 years ago

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

Pepsi [2]

Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

$a_r=\omega^2R$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2$

Now, tangential acceleration is given by the formula:

$a_t=R\alpha$

Plug in the given values and solve for $a_t$ . This gives,

$a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2$

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

$a_{Total}=\sqrt{(a_r)^2+(a_t)^2}$

Plug in the given values and solve for total acceleration, $a_{Total}$ . This gives,

$a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2$

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

$\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°$

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0

3 years ago