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BARSIC [14]
2 years ago
10

B. A car is moving 4.0 m/s to the right. The car begins to accelerate at a rate of 1.5 m/s/s, to the right. After

Physics
1 answer:
Maurinko [17]2 years ago
7 0
There you go, hope that helps.

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What experimental evidence led to the development of this atomic model from the one before it?
Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0
2 years ago
Read 2 more answers
Solve the question that follows using the equation for the conversion of Celsius to Fahrenheit. F=95(C)+32 On February 9, 1934,
SVEN [57.7K]

Answer:

Buffalo, NY

Explanation:

Temperature in Buffalo, NY = -29°C

In order to compare the temperatures we need to convert them to the same scale.

F=\dfrac{9}{5}C+32\\\Rightarrow F=\dfrac{9}{5}\times -29+32\\\Rightarrow F=-20.2

So, the temperature in Buffalo, NY was -20.2°F and the temperature in Anchorage, AL was 19°F.

Hence, it was colder in Buffalo, NY than in Anchorage, AL.

3 0
2 years ago
An ambulance is traveling north at 55.9 m/s, approaching a car that is also traveling north at 28.4 m/s. The ambulance driver he
wlad13 [49]

Answer:

915 Hz

Explanation:

The observed frequency from a sound source is given as

f₀ = f [(v + v₀)/(v+vₛ)]

where

f₀ = observed frequency of the sound by the observer = ?

f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

v₀ = velocity of the observer = 28.4 m/s

f₀ = 983 [(343+28.4)/(343+55.9)]

f₀ = 915.2 Hz = 915 Hz

6 0
3 years ago
A bubble at the bottom of lake rises to surface within 10.0 seconds with an acceleration of 10.0 meters per second. What is the
amm1812
If the bubble travels 10 meters per second and it takes 10 seconds, then just multiply the distance per second by the total seconds to get the total depth.
10 • 10 = 100
The lake is 100 meters deep.

Think of it this way to clarify the answer:
It takes a bubble traveling at a speed of 10 meters per second 10 seconds to travel 100 meters.
5 0
2 years ago
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
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