In the formula Q = m × c ×Δ T, which symbol represents specific heat? c m Q Δ T

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

EASY BRAINLIEST PLEASE HELP!!

jeka94

Answer:

Solution given:

frequency[f]=60,500,000Hz

velocity[V]=300,000,000m/s

wave length=?

we have

wave length= $\frac{V}{f}$

= $\frac{300,000,000}{60,500,000}$

= $\frac{3000}{605}$ =4.96 m

Option A.4.96m

3 0

3 years ago

Read 2 more answers

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

Svetach [21]

Answer:

The centripetal acceleration of the car will be 12.32 m/s² .

Explanation:

Given that

radius ,R= 57 m

Velocity , V=26.5 m/s

We know that centripetal acceleration given as follows

$a_c=\dfrac{V^2}{R}$

Now by putting the values in the above equation we get

$a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2$

Therefore the centripetal acceleration of the car will be 12.32 m/s² .

7 0

3 years ago

An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the s

Eddi Din [679]

Answer:

The correct answer is option 'a' 'The momentum is always conserved while as the kinetic energy may be conserved'

Explanation:

The conservation of momentum is a basic principle in nature which is always valid in an collision between 'n' number of objects if there are no external forces on the system. It is valid for both the cases weather the collision is head on or glancing or weather the object is elastic or inelastic.

The energy is only conserved in a collision that occurs on a friction less surface and the objects are purely elastic. Since in the given question it is mentioned that only the surface is friction less and no information is provided regarding the nature of the objects weather they are elastic or not hence we cannot conclusively come to any conclusion regarding the conservation of kinetic energy as the objects may be inelastic.

3 0

4 years ago

Students work together during an experiment about Newton’s laws. The students use a setup that consists of a cart of known mass

BartSMP [9]

Answer:

The tension will double and the acceleration will remain the same.

Explanation:

In order to solve this problem we must start by doing a drawing of the situation and by drawing free bodies diagrams for both elements. (See attached picture)

So let's analyze the first free body diagram. We can suppose the friction between the cart and the horizontal surface is zero and since we only care about the horizontal movement, we only take the horizontal forces into account. So we do a sum of forces:

$\sum F_{1}=m_{1}a$

since the only horizontal force for the first mass is the tension, we can say that:

$T=m_{1}a$

Now we can analyze the second mass. We will suppose the positive direction of the movement will be in the downwards direction, so we do a sum of forces:

$\sum F_{2}=m_{2}a$