Map.

The tires of a car make 75 revolutions as the car reduces its speed uniformly from 95 km/hr to 55 km.hr. the tires have a diamet

sveta [45]

<span>95 km/h = 26.39 m/s (95000m/3600 secs) 55 km/h = 15.28 m/s (55000m/3600 secs) 75 revolutions = 75 x 2pi = 471.23 radians radius = 0.80/2 = 0.40m v/r = omega (rad/s) 26.39/0.40 = 65.97 rad/s 15.28/0.40 = 38.20 rad/s s/((vi + vf)/2) = t 471.23 /((65.97 + 38.20)/2) = 9.04 secs (vf - vi)/t = a (38.20 - 65.97)/9.04 = -3.0719 The angular acceleration of the tires = -3.0719 rad/s^2 Time is required for it to stop (0 - 38.20)/ -3.0719 = 12.43 secs How far does it go? 65.97 - 38.20 = 27.77 M</span>

7 0

3 years ago

a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.

inessss [21]

Answer:

B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

B = $- \frac{P}{ \frac{\Delta V}{V} }$

The negative fate is for the module to be positive since the volume change is negative

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

B = $- \frac{10^4}{ \frac{-0.20}{5000} }$ = $\frac{10^4}{4 \ 10^{-5} }$

B = 2.5 10⁸ Pa

4 0

3 years ago

Which tuning forks will give rise to a beat frequency of 20 Hz when sounded together with a 240 Hz tuning fork?

Lubov Fominskaja [6]

Use the concept of beat frequency to find the applicable final freqeuncy for 20Hz beat frequency.

Beat can be defined as 'the interference pattern between two sounds of slightly different frequencies0

The expression for beat frequency is given as

$f_{beat} = |f_1-f_2|$

Where,

$f_2 =$ Final frequency

$f_1 =$ Initial frequency

The beat frequency for us is 25Hz and the initial frequency is 240Hz, then

$20= |f_2-240|$

Being an absolute value, two values are possible, both in addition and subtraction:

$f_2 = 240 \pm 20$

The two possible values are

$f_2 = 220Hz$

$f_2 = 260Hz$

3 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago

What is the relationship between the normal force and weight

GalinKa [24]

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

4 0

3 years ago