I’m having trouble with this

A car that weighs 1.0 x 10^4 N is initially moving at a speed of 38 km/h when the brakes are applied and the car is brought to a

hram777 [196]

Answer:

Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

$v^2=u^2+2as$

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr = $\frac{38\times 1000}{3600}=10.56m/s$

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

$0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2$

Now the force can be obtained using newton's second law as

$Force=\frac{Weight}{g}\times a$

Applying values we get

$Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons$

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

$v=u+at$ with symbols having the same meanings

Applying values we get

$0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds$

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0

3 years ago

A rectangular block of copper metal has a mass of 100 g. The dimensions of the block are 2 cm by 2 cm by 2 cm. From this data, w

lina2011 [118]

Can I have some more detail

6 0

3 years ago

Read 2 more answers

Energy can be described in many ways. From the list below, what is the best definition of energy?

max2010maxim [7]

the strength and vitality required for sustained physical or mental activity.

"changes in the levels of vitamins can affect energy and well-being"

synonyms: vitality, vigor, life, liveliness, animation, vivacity, spirit, spiritedness, verve, enthusiasm, zest, vibrancy, spark, sparkle, effervescence, ebullience, exuberance, buoyancy, sprightliness; More

2.

power derived from the utilization of physical or chemical resources, especially to provide light and heat or to work machines.

Have a great day ~Matya

8 0

3 years ago

Read 2 more answers

6) A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10−3 s

Romashka [77]

Answer:

a) L = 2.10x10⁴⁰ kg*m²/s

b) τ = 1.12x10²⁴ N.m

Explanation:

a) The angular momentum (L) of the pulsar can be calculated using the following equation:

$L = I \omega$

Where:

I: inertia momentum

ω: angular velocity

First we need to calculate ω and I. The angular velocity can be calculated as follows:

$\omega = \frac{2 \pi}{T}$

Where:

T: is the period = 33.5x10⁻³ s

$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s$

The inertia moment of the pulsar can be calculated using the following relation:

$I = \frac{2}{5}mr^{2}$

Where:

m: is the mass of the pulsar = 2.8x10³⁰ kg

r: is the radius = 10.0 km

$I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2}$

Now, the angular momentum of the pulsar is:

$L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1}$

b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:

$\tau = I*\alpha$

Where:

α: is the angular acceleration = 10⁻¹⁴ rad/s²

$\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m$

I hope it helps you!

5 0

3 years ago

I'm completely stumped as to how to do this.

worty [1.4K]

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force $\vec{\textbf{R}}$ and F1 are completely along the x-axis, we can conclude that