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goldfiish [28.3K]
3 years ago
14

How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t

he field? He is travelling at 4 m/s
Physics
1 answer:
kkurt [141]3 years ago
4 0
Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s

t=12s
You might be interested in
A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surfa
Burka [1]

Answer:

(a) = 0 N

(b) = 2.4 N

Explanation:

given

box of banana weight = 40.0 N

coefficient of static friction  μ = 0.40

coefficient of kinetic friction = 0.20

a).  when no horizontal force is applied .

according to Newton 's third law of motion If there is no force applied to the box,so the frictional  force exerted is 0 N

b) magnitude of friction force

box and the box is initially at rest

    friction force =.Static friction coefficient × weight of the box

      friction force = 0.40 × 6

       friction force =  2.4 N

8 0
2 years ago
Suppose a scoentist was able to construct a barometer with a liquid being denser than mercury , then how high would the liquid r
ki77a [65]

Answer:

the liquid has less height than the mercury

      h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

Explanation:

The pressure as a function of the height is given by

        P = ρ g h

where ρ is the density of the liquid, g the acceleration of gravity and h the height reached by the column of the liquid

In that case they say that the pressure is the standard one that is P = 1.01 10⁵ Pa = 760 mmHg

The first way to give the pressure is in SI units and the second way is the height that the mercury column reaches

In the case of building a barometer with a liquid that has a density greater than that of mercury

            ρ_liquid > ρ_Hg

             

the pressure

              P =ρ_lquid g h_liquid

if we have the same pressure

            ρ_{Hg} g h_{Hg} = ρ_{liquid}  g h_{liquid}

            h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

therefore the liquid has less height than the mercury

7 0
3 years ago
The emission spectrum of iodine is shown below.<br> Which is the absorption spectrum?
garik1379 [7]
Sorry Sorry Sorry Sorry Sorry Sorry Sorry Sorry
6 0
3 years ago
Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0
3 years ago
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