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ella [17]
3 years ago
12

1. What are vaccines and why are they important for people's health​

Physics
1 answer:
Evgen [1.6K]3 years ago
8 0
To help fight off disease and viruses
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An ant is on the pendulum bob of the clock. What would be the displacement in one time period?
Dennis_Churaev [7]

At the end of one full time period, the ant has returned to where it was at the beginning of the time period. Its displacement is <em>zero</em>.

6 0
3 years ago
A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc
katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

V_{ds}=V_{river}+V{boat}

V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

V_{us}=V_{boat}-V{river}

V_{us}=5.60km/hr-2.80km/hr=2.80km/hr

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

4 0
4 years ago
Read 2 more answers
A 30kg mass is brought from the earth to the moon what is the weight on the earth
ozzi

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

2

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

8 0
2 years ago
How do ocean currents on the water's surface transfer energy?
Anit [1.1K]
Exactly what the first person said. Hope this helps!
6 0
3 years ago
Hoop rolling up an inclined plane A hollow cylinder (or hoop) is rolling along a horizontal surface with speed v = 3.3 m/s when
Naya [18.7K]

Answer:

Explanation:

For rolling up or down an incline plane , the acceleration or deceleration of the rolling body is given by the following expression

a = g sinθ / (1 + k²/r² )

where k is radius of gyration of rolling body and θ is angle of inclination

a = g sin15 / ( 1 + 1 ) [ for hoop k = r ]

a = 9.8 x .2588 / 2

= 1.268 m / s²

a )

Let s be the distance up to which it goes

v² = u² - 2as

0 = 3.3² - 2 x 1.268 s

s = 4.3 m

b ) Let time in going up be t₁

v = u - at₁

0 = 3.3 - 1.268 t₁

t₁ = 2.6 s

Time in going down t₂

s = 1/2 a t₂²

4.3 = .5 x 1.268 t₂²

t₂ = 2.60

Total time

= t₁ +t₂

= 2.6 + 2.6

= 5.2 s

4 0
4 years ago
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