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3 years ago

60 POINT!!!!!!

Physics

2 answers:

Paladinen [302]3 years ago

7 0

Answer:

h and I have been working on the sea for the last few days

Wittaler [7]3 years ago

4 0

Answer:

Samples 2 and 4 are the same liquid

Explanation:

Samples 2 and 4 should be the same liquid because they have the same density as well as the same boiling point. The mass and volumes given do not matter in this case, however they could be used to calculate the density which was already given.

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A rock dropped into a pond produces a wave that takes 11.3 s to reach the opposite shore, 26.5 m away. the distance between cons

Kay [80]

The wave takes 11.3 s to cover a distance of 26.5 m, so its speed is:
$v= \frac{S}{t}= \frac{26.5 m}{11.3 s}=2.35 m/s$

The distance between two consecutive crests is 3 m, and this corresponds to the wavelength of the wave. To find its frequency, we can use the relationship between the speed v, the wavelength $\lambda$ and the frequency f:
$f= \frac{v}{\lambda}= \frac{2.35 m/s}{3 m}=0.78 Hz$

6 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

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

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

find the rate of positive acceleration of an automobile which went from a complete stop to a velocity of 30 meters per second in

maks197457 [2]

Answer:

3 m/s^2

Explanation:

acceleration= Change in velocity/time

= 30-0 / 10

= 30/10

=3 m/s^2

3 0

3 years ago

A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the

Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

v = u + at

0 = 19.09 - 9.81 t

t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

S = ut + 0.5 at²

H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

Maximum height of the ball = 18.57 m

6 0

3 years ago

An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?

NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object = 49N

Force applied = 12N

Unknown:

Acceleration of object = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

Acceleration = $\frac{12}{49}$ = 0.25m/s²

3 0

2 years ago