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3 years ago

5 sentences about the mouth

Physics

2 answers:

AlladinOne [14]3 years ago

7 0

Awww im too late but im going to wait and see if he has it

Likurg_2 [28]3 years ago

4 0

Answer:

mouth is to eat
mouth helps us to drink water
mouth helps us to breath
mouth is in head
mouth contain tongue which helps us to taste

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A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves

Len [333]

Answer:

the velocity in all the cases will be same.

Explanation:

given,

girl throws a stone from the bridge

air is friction less

we have to find from the given cases in which case the velocity of stone will be greatest.

According to Work energy theorem work done by the sum of all the force is equal to kinetic energy.

As the air is frictionless hence the speed depend upon the height from which the stone is thrown.

height in all the cases is same.

so, the velocity in all the cases will be same.

3 0

2 years ago

Read 2 more answers

A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot

antiseptic1488 [7]

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

$v_f^2=v_i^2+2as$

Here,

$v_f$ = Final velocity

$v_i$ = Initial velocity

a = Acceleration

s = Distance

Replacing,

$(35)^2 = v_i^2+2(1.5)(392)$

$v_i = 7m/s$

Using the conservation of energy for kinetic energy we have,

$KE = \frac{1}{2}mv_i^2$

$KE = \frac{1}{2}(1302)(7)^2$

$KE = 31900J$

Therefore the kinetic energy of the car is 31900J

5 0

3 years ago

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just

bagirrra123 [75]

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

5 0

3 years ago

A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu

Anettt [7]

Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

$a=\mu g$

and maximum acceleration during SHM is

$a=A\omega ^2$

$a=A(2\pi f)^2$

$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

$a=5.296\ m/s^2$

$\mu =\frac{a}{g}$

$\mu =\frac{5.296}{9.8}=0.54$

6 0

3 years ago

Read 2 more answers

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

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

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

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

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

3 years ago

5 sentences about the mouth​

5 sentences about the mouth