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1 year ago

Which of these is a push or a pull? Acceleration Force Mass Inertia

Physics

1 answer:

Fantom [35]1 year ago

5 0

Answer:

the answer is force . force is applied as a push or pull

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The length of the mercury thread is found to be 4cm and 24cm at ice point and steam point respectively on an ungraduated thermom

BabaBlast [244]

Answer:

The difference between ice and steam in Celsius (Centigrade) is 100 deg.

So the difference between and 4 cm and 24 cm of the thread corresponds to 100 deg C.

So 8 cm is 4 cm greater than the ice point

4 cm / 20 cm = 1/5 since the steam point and the ice point are 20 cm apart

Then 1/5 * 100 deg C = 20 deg C the requested temperature

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2 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

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2 years ago

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.5 cm/s on an otherwise placid lake.

nata0808 [166]

Find the given attachment

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2 years ago

A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.

enyata [817]

Answer:

The induced emf between two end is $34.02 \times 10^{-5}$ V

Explanation:

Given:

Length of rod $l = 1$ m

Height $h = 1.23$ m

Magnetic field $B = 6.93 \times 10^{-5}$ T

For finding induced emf,

$\epsilon = Blv$

Where $v =$ velocity of rod,

For finding the velocity of rod.

From kinematics equation,

$v^{2} = v_{o} ^{2} + 2gh$

Where $v_{o} =$ initial velocity, $g = 9.8 \frac{m}{s^{2} }$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.8 \times 1.23}$

$v = 4.91$ $\frac{m}{s}$

Put the velocity in above equation,

$\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91$

$\epsilon = 34.02 \times 10^{-5}$ V

Therefore, the induced emf between two end is $34.02 \times 10^{-5}$ V

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2 years ago

How does biodiversity help sustain populations of organisms?

tia_tia [17]

Biodiversity helps sustain populations of organisms by making it where if one organism dies it does not go extinct because there is many different type of that organism left

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2 years ago