How much heat must be removed

In Speed Study Number 1, we looked at two cars traveling the same distance at different speeds on city streets. Car "A" traveled

Ganezh [65]

Answer:

230.4 s

Explanation:

The speed of car A is

$v_A = 35 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car A is

$t_A = \frac{d}{v_A}=\frac{10 mi}{35 mi/h}=0.286 h$

The speed of car B is

$v_B = 45 mi/h$

and the distance travelled is

$d = 10 mi$

so the time taken for car B is

$t_B = \frac{d}{v_B}=\frac{10 mi}{45 mi/h}=0.222 h$

So the difference in time is

$\Delta t = t_A - t_B = 0.286 h -0.222 h=0.064 h$

Which corresponds to

$\Delta t = 0.064 h \cdot 3600 s/h = 230.4 s$

so car B arrived 230.4 s before car A.

5 0

3 years ago

what will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.30 meters in the downwar

Anna35 [415]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices the can be found elsewhere:

A.) 14 newtons upward
B.) 45 newtons upward 
C.) 67 newtons upward 
D.) 130 newtons upward 
E.) 150 newtons upward

The answer is A.) 14 newtons upward

7 0

3 years ago

Two blocks of clay, one of mass 1.00 kg and one of mass 7.00 kg, undergo a completely inelastic collision. Before the collision

RideAnS [48]

Answer:8 m/s

Explanation:

Given

$m_1=1 kg$

$m_2=7 kg$

kinetic Energy of $m_1=32 J$

initially $m_2$ is at rest and let say $m_1$ is moving with velocity u

kinetic Energy of $m_1 is =\frac{m_1u^2}{2}=32$

$u^2=64$

$u=8 m/s$

In Completely inelastic collision both mass stick together and move with common velocity

Suppose v is the common velocity

$m_1u+0=(m_1+m_2)v$

$v=\frac{1\times 8}{1+7}=1 m/s$

therefore Final velocity with which both blocks moves is 1 m/s

7 0

3 years ago

2 Which is true of a parallel circuit?

mars1129 [50]

Answer:

fxb

c

Explanation:bfffffffff

8 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

3 years ago

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