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Svetlanka [38]
3 years ago
15

Two objects, labeled A and B, are the same size. Object A has a density of 1.21 g/cm3. Object B has a density of 1.37 g/cm3. Bot

h are placed in a beaker of water. Which will float higher in the water?
Physics
2 answers:
Sedaia [141]3 years ago
4 0

Neither object will float in water. 

They both have densities greater than 1.0 g/cm³,
so both will sink to the bottom of the beaker.

Like rocks.

deff fn [24]3 years ago
4 0

Answer:

None of the object float in water

Explanation:

When an object float in water than we can say that buoyancy force due to water will counter balance the weight of the object

As we can say that

W_g = F_b

so here if object is just floating or we can say that if object is in the state of sinking then we have

\rho_{object} V g = \rho_{water} Vg

so we have

\rho_{object} = \rho_{water}

so from above relation we can say that

i)\rho_{object} > \rho_{water} then it will sink inside water

ii)\rho_{object} < \rho_{water} then it will float in water

since we know density of water is 1 gm/cm^3 so here given both objects are having density more than this

so both the objects will sink inside water

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Starting from rest, a dragster travels a straight 1/4 mi racetrack in 7.10 s with constant acceleration. What is its velocity wh
Gennadij [26K]

268.6567 mph  is its velocity when it crosses the finish line

d=(v1+v2 /2) x t

.25=(0+v2 /2) x 6.7/3600 hours

900=v2/2 x 6.7

v2=268.6567 mph as the speed with which the dragster crosses the finish

<h3>When acceleration is not zero, can speed remain constant?</h3>

The answer is that an accelerated motion can have a constant speed. Consider a particle travelling uniformly around a circle; it experiences acceleration since the motion's direction is changing, but it maintains a constant speed along the tangential axis throughout the motion.

Acceleration is the frequency of a change in velocity. Acceleration is a vector with magnitude and direction, much as velocity. For instance, if a car is moving in a straight path and speeding up, it is said to have forward (positive) acceleration, and if it is slowing down, it is said to have backward (negative) acceleration.

Learn more about velocity refer

brainly.com/question/24681896

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5 0
1 year ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

The range is 35.35 m

7 0
3 years ago
Suzie skydiver, who weighs 500 n, reaches terminal velocity of 90 km/h. the air resistance on suzie is then
masha68 [24]
<span>Weight of the skydiver m = 500 N
 Terminal velocity V = 90 km/h
 Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N</span>
7 0
3 years ago
Read 2 more answers
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

And the wave speed is given by:

v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0
3 years ago
A hockey player skates with an average speed of 8.25 m/s. The distance the player will travel in 5.60 s is __________ .
OverLord2011 [107]
<h2>Greetings!</h2>

To find this value, you need to remember the speed formula:

3 = 6 / 2

Speed = distance ÷ time

Rearrange to make distance the subject:

Distance = speed * time

Simply plug these values into this:

5.6 * 8.25 = 46.2

<h3>So the player will travel 46.2 metres!</h3>
<h2>Hope this helps!</h2>
7 0
2 years ago
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