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3 years ago

A car goes from rest to a speed of 90 km/h in 10 seconds . What is the car's acceleration in m/s²

1 answer:

7 0

First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s

Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!

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Verizon [17]

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4 0

3 years ago

A certain computer chip that has dimensions of 3.67 cm and 2.93 cm contains 3.5 million transistors. If the transistors are squa

Evgesh-ka [11]

Answer:

1.56 × 10^-3 cm.

Explanation:

So, we are given the following parameters from the question above;

Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.

Step one: find the area of the computer chip.

Therefore, Area = Length × breadth.

Area = 3.67 cm × 2.93 cm.

Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.

Step two: find the area of one transistor

The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).

Hence;

The area of one transistor= 10.7531/4.4 × 10^6.

The area of one transistor= 2.44 × 10^-6 cm^2.

=> Note that We have our transistors as square, therefore;

The maximum dimension = √ (2.44 × 10^-6) cm^2.

The maximum dimension= 1.56 × 10^-3 cm.

7 0

3 years ago

A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find

pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0=4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

$I=1.41 A$ (current in the wire)

$B=5.61\mu T=5.61\cdot 10^{-6} T$ (strength of magnetic field)

Solving for r, we find the distance from the wire:

$r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m$

4 0

3 years ago

Pls answer 50 points for an answer!!!!

lyudmila [28]

ummmn hi i dont know lol

5 0

3 years ago

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done

blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

$\Rightarrow a=\frac{F}{m}$

Hence acceleration of the block is given by:

$\Rightarrow a=\frac{25}{2}=12.5m/s^2$

Displacement of the object is given by:

$\Rightarrow S=ut+\frac{1}{2}at^2$

Substitute the values

$\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2$

$\Rightarrow S=56.25 m$

Now work done is given by:

W=F.S

W = 25×56.25

W =1406.25 J

3 0

3 years ago