Is my answer correct plz explain i don't understand it.

A car is traveling at 21m/s. It accelerates at 1.4m/s2 for 11s. How fast is the car moving after the acceleration?

nata0808 [166]

Answer:

v=u+at

v=21+1.4(11)

v=21+15.4

v=36.4

3 0

3 years ago

Identify three situations in which convection occurs

hodyreva [135]

There are three types of heat conduction through substances. These are named as conduction,convection and radiation .here we have been given convection.

Convection is the type of mode of conduction of heat in which heat will flow though a liquid and gases due to the direct physical movement of particles.In this process the hotter particles will go upward as they become lighter and cooler,heavier particles come downward which after being heated up go upward .Hence a convectional current is formed for which whole of the liquid or gas gets heated.

There are different life examples of convection.

One may take an simple example of water in a container.The water molecules which are present bottom part of the container will be heated up first and go upward.The upper particles will come downward and they will constitute a convectional current.

Another life example is the flow of wind from one region to another region.the air at hotter region will become lighter and goes upward and the wind starts flowing from cold region region in order to occupy the vacant space.

Another example is the hot air balloon rising up..It is also another example of convection of heat

3 0

3 years ago

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A stuntman drives a car with a mass of 1500 kg on a drawbridge. The car accelerates with a constant force of 10,000 N. While he

klemol [59]

The cars new accelleration is 9,000 N.

5 0

3 years ago

The force between two point-charges of 0.040 C and 0.060 C separated by certain distance is 2.2 x 10^5 N. what is the distance b

Gala2k [10]

Answer:

im in hardvard dont cheat and get good grades and become a graduate

it is c

Explanation:

8 0

3 years ago

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A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo

statuscvo [17]

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase (or decrease) in length of the object.

In our problem, $L_0 = 3.0 m$ is the initial length of the column, $E=3.0 \cdot 10^{10}N/m^2$ is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
$r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m$
and the cross-sectional area is
$A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2$
The force applied to the column is the weight of the load:
$W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N$

Now we have everything to calculate the compression of the column:
$\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m$
So, the column compresses by 1.83 millimeters.

3 0

3 years ago

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