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3 years ago

21. A toy car starts from rest and accelerates at 1.50 m/s’ [E] for 5.25 s. What is the final

Physics

1 answer:

yulyashka [42]3 years ago

7 0

Answer: The final Velocity, V, of the car is 7.9m/s

Explanation:

Given the following :

Toy car starts from rest, therefore,

Initial Velocity (u) = 0

Acceleration (a) = 1.5m/s^2 E for time, t = 5.25s

Final velocity (V) of the car =?

Using the motion equation:

v = u + at

Where v = final Velocity

u = Initial Velocity

a = acceleration

t = time

v = 0 + 1.5(5.25)

v = 0 + 7.875

v = 7.875m/s

v = 7.9m/s

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Igneous rocks are formed from _____.

Leni [432]

Answer:

magma

Explanation:

actually cooling and solidification of magma

7 0

3 years ago

If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

8 0

4 years ago

The diver on the diving board is 10 meters high and has a mass of 0.050kg, what is his GPE?

sveta [45]

Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 10 × 9.8 × 0.05

We have the final answer as

Hope this helps you

6 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago

Hurricane sandy produced some of the greatest destruction along the new jersey coast in communities situated along narrow strips

FrozenT [24]

The correct answer to fill in the blank would be:

“a barrier island”

Barrier islands are coastal landforms and a category of dune system that are remarkably even or lumpy areas of sand that was formed by wave and tidal actions that are parallel to the mainland coast. Due to this feature, there are no enough sand blockades to minimize the destruction by Hurricane Sandy.

4 0

3 years ago