Ask question

Ask question

All categories

3 years ago

13

After a while, the car started to go around a long bend, still maintaining its constant speed of 55 miles per hour. Is there a n

et (unbalanced) force acting on the car? 1. No; the speed of the car does not change. 2. Cannot be determined 3. No; the velocity of the car does not change. 4. Yes; its direction is not the same as the velocity. 5. Yes; its direction is the same as the velocity.

1 answer:

sukhopar [10]3 years ago

6 0

Answer:

4) True. The change of direction needs an unbalanced force

Explanation:

Let us propose the resolution of the problem using Newton's second law.

F = m a

As the car is spinning the acceleration is centripetal

a = v2.r

F = m v2 / r

We can see that as the velocity of a vector even if its module does not change, the change of direction requires an external force.

Now we can analyze the statement if they are true or false

1) and 3) False, even when the speed changes, the direction changes

2) False with the speed change can be determined

4) True. The change of direction needs an unbalanced force

5) False are different things. the direction is where it is going and the speed is the magnitude of the vector

You might be interested in

Choose the best real world exzample of convection:

spin [16.1K]

The correct answer is A

3 0

3 years ago

Read 2 more answers

Acceleration question plz help

andrezito [222]

For #5 It's helpful to draw a free body diagram so you know which way the forces are acting on the block.

the weight mg is acting downwards, and you need to find the vertical and horizontal components of mg using sin and cosine. so do 15x9.8xsin40 which is the force. Assuming no friction, this is the only force acting on the block, as the forces on the vertical plane cancel out i.e the normal force and weight of the block.

after, just do F=ma And since you know F and m, solve for a.

3 0

3 years ago

Peak wavelength of light coming from a star with a temp of 4300k

PSYCHO15rus [73]

from Wein's law

$\frac{k}{ \lambda_{max} } = T \\ \lambda_{max} = \frac{k}{T} \\ \lambda_{max} = \frac{3.898 \times {10}^{ - 3} }{4300} \\ = 9.065 \times {10}^{ - 7}$

3 0

3 years ago

The current theory of the structure of the

IRISSAK [1]

1) The mass of the continent is $3.3\cdot 10^{21} kg$

2) The kinetic energy of the continent is 624 J

3) The speed of the jogger must be 4 m/s

Explanation:

1)

We start by finding the volume of the continent. We have:

$L = 5850 km = 5.85\cdot 10^6 m$ is the side

$t = 35 km = 3.5\cdot 10^4 m$ is the depth

So the volume is

$V=L^2 t = (5.85\cdot 10^6)^2 (3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that its density is

$d=2750 kg/m^3$

Therefore, we can find the mass by multiplying volume by density:

$m=dV=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by:

$K=\frac{1}{2}mv^2$

where

$m=3.3\cdot 10^{21} kg$ is its mass

v = 3.2 cm/year is the speed

We have to convert the speed into m/s. We have:

3.2 cm = 0.032 m

$1 year = 1(365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is:

$v=\frac{0.032 m}{3.15 \cdot 10^7 s}=1.02\cdot 10^{-9} m/s$

So, we can now find the kinetic energy:

$K=\frac{1}{2}(1.20\cdot 10^{21})(1.02\cdot 10^{-9})^2=624 J$

3)

Here we have a jogger of mass

m = 78 kg

And the jogger has the same kinetic energy of the continent, so

K = 624 J

The kinetic energy of the jogger is given by

$K=\frac{1}{2}mv^2$

where v is the speed of the jogger.

Solving for v, we find the speed that the jogger must have:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(624)}{78}}=4 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

A baseball pitcher throws a ball at 97.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it

mario62 [17]

Answer:

Explanation:

1 mile=5280 ft

1 hour=3600 seconds

Changing speed from mi/h to ft/s

$97 mi/h \times \frac {5280}{3600}=142.2666667 ft/s\approx 142.27 ft/s$

Time of flight, $t=\frac {distance}{speed}=\frac {60.5}{142.27}=0.425257732 s\approx 0.425 s$

From fundamental kinematics equations

$h=0.5gt^{2}$ where g is acceleration due to gravity whose value is taken as 32.2 ft/s2 and h is the distance

By substitution

$h=0.5\times 32.2\times 0.425^{2}=2.9080625 m\approx 2.91 m$

7 0

3 years ago