What burns quicker, white or colored candles?

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

7 0

3 years ago

the volume of a water tank is 5m×4m×2m. If the tank is half filled with water. calculate the pressure exerted at the bottom of t

Natali5045456 [20]

Answer:

10⁴ Pa

Explanation:

Volume = 5m * 4m * 2m

When the tank is half filled ,

Volume = 5m * 4m * 1m = 20m³

Also we know that ,

Density of water = 10³ kg/ m³

$\longrightarrow$ Density = Mass/Volume

$\longrightarrow$ 10³ kg/m³ = m/20m³

$\longrightarrow$ m = 2 * 10⁴ kg

So that ,

$\longrightarrow$ Weight = mg

$\longrightarrow$ F = 2*10⁴ × 10 N

$\longrightarrow$ F = 2 * 10⁵ N

And ,

$\longrightarrow$ Pressure = Force/Area

$\longrightarrow$ P = 2 *10⁵/ 5 * 4 Pa

$\longrightarrow$ P = 10⁴ Pa

5 0

2 years ago

Examine the nuclear reaction: Superscript 1 subscript 1 H plus superscript 1 subscript 0 n right arrow superscript 2 subscript 1

AysviL [449]

Answer:

C. A change has occurred in the nucleus.

Explanation:

6 0

2 years ago

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What is the equation to find an angle of projectile

Ksju [112]

I do not recall the answer to this question

3 0

2 years ago

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

Alex777 [14]

It’s the type of eclipse that occurred when the moon passes between the sun and earth, and when the moon fully or partially blocks the sun.

7 0

2 years ago