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matrenka [14]
1 year ago
7

Two identical grasshoppers jump into the air with the same initial speed and experience no air resistance. Grasshopper A goes st

raight up, but grasshopper B goes up at a 66° angle above the horizontal. Which of the following statements about these grasshoppers are correct? (There could be more than one correct choice.
Choices below:

a.) At their highest point, both of them have the same amount of mechanical energy.
b.) At their highest point, grasshopper B is moving faster than grasshopper A.
c.) At their highest point, both of them have the same amount of kinetic energy.
d.) At their highest point, both of them have the same amount of gravitational potential energy.
e.)At their highest point, grasshopper A has more gravitational potential energy than grasshopper B.
Physics
1 answer:
avanturin [10]1 year ago
7 0

Answer:

a.) At their highest point, both of them have the same amount of mechanical energy.

Explanation:

Grasshopper A jumps upwards with speed v while grasshopper B jumps at an angle of 66 degree with horizontal at same speed

so vertical speed of B is given

v_y = v sin66 = 0.91v

v_x = v cos66 = 0.41v

So when both the grasshopper will jump upwards they both will attain different maximum heights

Since initial speed of both grasshopper is same so their mechanical energy is always conserved

So mechanical energy of grasshopper A is given as

U = \frac{1}{2}mv^2

Mechanical energy of grasshopper B is given as

U = \frac{1}{2}m(vcos^2\theta) + \frac{1}{2}m(vsin\theta)^2

U = \frac{1}{2}mv^2

So we have

a.) At their highest point, both of them have the same amount of mechanical energy.

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Sea breezes that occur near the shore are attributed to a difference between land and water with respect to what property?
ddd [48]

Answer:

a. mass density

Explanation:

<em>Land and sea breeze that occur near the shore are due to the variation of mass density of air with change in temperature.</em>

  • When the air gets heated it becomes rarer in density and thus rises up in the atmosphere and its space is occupied by a cooler and denser air that flows to the place.

<em>During the day the land is warmer than the sea so the sea breeze blows and during the night the water bodies are warmer than the land so the land breeze blows.</em>

7 0
2 years ago
A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
Read 2 more answers
g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb
Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber  = v x 1 x g

Buoyant force on lead =  .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g  = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115  / 33.5

= .36  

= 36 %

4 0
2 years ago
You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel
Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

T = 2 \pi \sqrt{\frac{m}{k}}

squaring on both sides,

T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

y=T^2=0.0569x+0.0010

the hanging mass M is x here, so comparing the equation we know that

\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m

b) In order to find the mass of the spring we make similar process, so comparing,

\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g

3 0
2 years ago
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas i
asambeis [7]

This question is incomplete, the complete question is;

The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.

"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"

Answer:

the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

Explanation:

Given that;

P₁ = 1.00 atm

P₂ = ?

V₁ = 1 L

V₂ = 1.60 L

the temperature of the gas is kept constant

we know that;

P₁V₁ = P₂V₂

so we substitute

1 × 1 = P₂ × 1.60

P₂ = 1 / 1.60

P₂ = 0.625 atm

Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant

5 0
2 years ago
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