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3 years ago

PLZ SOMEONEE HELPP I’LL MARK BRANLIESTTTT

Physics

1 answer:

7nadin3 [17]3 years ago

8 0

Answer:

I'm pretty sure it's 37.5 joules of energy

Explanation:

hope this helps!

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Erase all the trajectories, and fire the pumpkin vertically again with an initial speed of 14 m/s. As you found earlier, the max

yanalaym [24]

Answer:

$\theta=39.49^{\circ}$

Explanation:

Maximum height of the pumpkin, $H_{max}=9.99\ m$

Initial speed, v = 22 m/s

We need to find the angle with which the pumpkin is fired. the maximum height of the projectile is given by :

$H_{max}=\dfrac{v^2\ sin^2\theta}{2g}$

On rearranging the above equation, to find the angle as :

$\theta=sin^{-1}(\dfrac{\sqrt{2gH_{max}}}{v})$

$\theta=sin^{-1}(\dfrac{\sqrt{2\times 9.8\times 9.99}}{22})$

$\theta=39.49^{\circ}$

So, the angle with which the pumpkin is fired is 39.49 degrees. Hence, this is the required solution.

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3 years ago

Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains

lakkis [162]

Given that the density of heptane is

$d_h=\frac{0.684g}{mL}$

The mass of heptane is

$m_h=31\text{ g}$

The density of water is

$d_w=\frac{1g}{mL}$

The mass of water is

$m_w=37\text{ g}$

The volume of heptane will be

$\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}$

The volume of water will be

$\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}$

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

$\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}$

The total volume of liquid in the cylinder will be 82.32 mL.

7 0

1 year ago

A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that

Readme [11.4K]

Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels l as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height h it reached, we can calculate l using the sin(α) function:

$sin(\alpha )=\frac{opposite side}{hypotenuse}\\sin(\alpha)=\frac{h}{l}\\l=\frac{h}{sin(\alpha)}$

Replacing:

$l=\frac{1000ft}{sin(\frac{\pi}{4})}$

Solving and roundin to the nearest foot:

$l=1414 ft$

3 0

3 years ago

For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr

leonid [27]

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

$W=m\times g$

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = $9.8m/s^2$

Putting values in above equation, we get:

$W=3kg\times 9.8m/s^2\\\\W=29.4N$

Hence, the weight of the object is 29.4 N

6 0

3 years ago

A football player with a mass of 88 kg and a speed of 2.0 m/s collides head-on with a player from the opposing team whose mass i

Ket [755]

Answer:

Speed of another player, v₂ = 1.47 m/s

Explanation:

It is given that,

Mass of football player, m₁ = 88 kg

Speed of player, v₁ = 2 m/s

Mass of player of opposing team, m₂ = 120 kg

The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :

$m_1v_1+m_2v_2=(m_1+m_2)V$

V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.

$v_2=-\dfrac{m_1v_1}{m_2}$

$v_2=-\dfrac{88\ kg\times 2\ m/s}{120\ kg}$

$v_2=-1.47\ m/s$

So, the speed of another player is 1.47 m/s. Hence, this is the required solution.

7 0

3 years ago