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3 years ago

Which is a characteristic of projectile motion?

Physics

2 answers:

professor190 [17]3 years ago

8 0

Answer:

Explanation:

The projectile motion is an example of two dimensions motion. It is a motion in two dimensions under the force of gravity.

When a projectile throws at any arbitrary angle other than 0 degree or 90 degree, it covers both horizontal and vertical distance.

The horizontal distance is called height and horizontal distance is called range.

The formula for the vertical height is given by

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

The formula for the horizontal range is given by

$R = \frac{u^{2}Sin 2\theta }{g}$

Ne4ueva [31]3 years ago

3 0

A projectile is an object upon which the only force acting is gravity. Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward or downward they are also moving horizontally.

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What is the momentum of an object that has a mass of 0.03 grams and a velocity of 1200 m/s2

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2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

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3 years ago

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The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a

Mkey [24]

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

$A=2(lb+bh+hl)$

Put the value into the formula

$A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)$

$A=0.85\ m^2$

Inner surface temperature of freezer

$T_{i}=-10°C=263\ K$

Outer surface temperature of freezer

$T_{o}=33+273=306\ K$

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

$q=\dfrac{kA}{L}(T_{o}-T_{i})$

$L=\dfrac{kA}{q}(T_{o}-T_{i})$

Put the value into the formula

$L=\dfrac{0.30\times0.85}{400}(306-263)$

$L=0.02741\ m$

Hence, Thickness of Styrofoam insulation is 0.02741 m.

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3 years ago