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2 years ago

9

The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the fl

ower in a mere 0.30 ms at an acceleration of 2.5 × 104 m. s2 If the acceleration is constant, what impulse is delivered to a pollen grain with a mass of 1.0 × 10−7g?

1 answer:

stellarik [79]2 years ago

3 0

Answer:

I = 7.5*10^-10 kg m/s

Explanation:

In order to calculate the impulse you first take into account the following formula:

$I=m\Delta v=m(v-v_o)$ (1)

m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg

v: final speed of the pollen grain = ?

vo: initial speed of the pollen grain = 0 m/s

Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:

$v=v_o+a t$ (2)

a: acceleration = 2.5*10^4 m/s^2

t: time = 0.30ms = 0.30*10^-3 s

$v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}$

Next, you replcae this value of v in the equation (1) and calculate the impulse:

$I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}$

The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s

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The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t

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Answer:

$g'=\frac{g__R}{4}$

Explanation:

Given:

gravitational field strength of moon at a distance R from its center, $g__R$

Distance of the satellite from the center of the moon, $h=2R$

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

$g'=g__{R}} \times \frac{R^2}{(2R)^2}$

$g'=\frac{g__R}{4}$

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

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2 years ago

Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.

Viefleur [7K]

Answer:900 feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

$v^2-u^2=2as$

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph $\approx 88.002 ft/s$

$v^2-u^2=2as$

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

8 0

2 years ago

Read 2 more answers

Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad

mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = $\rhoV$

V = Volume = $\dfrac{4}{3}\pi r^3$

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

$\rho$ = Density = 19 g/cm³

c = Speed of light = $3\times 10^8\ m/s$

A = Area = $\pi R^2$

I = Intensity = $\dfrac{P}{4\pi r^2}$

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

$F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}$

According to the question

$F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W$

The power required of the light source is 468449163762.0812 W

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2 years ago

A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. (a) What is the foca

xenn [34]

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance = 4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm$

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2$

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

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2 years ago

Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is

vaieri [72.5K]

1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.

angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.

8 0

2 years ago