What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

How much energy (in kWh) can you get from a 72 cell solar panel in one year?

Mumz [18]

Answer:72,000
Explanation:

4 0

3 years ago

Through what angle in degrees does a 33 rpm vinyl record turn in 0.23 s : A) 46° B) 26° C) 53° D) 33°

solniwko [45]

Answer:

(a) angle record will be 46°

Explanation:

We have given speed = 33 rpm

Time t = 0.23 sec

We know that 1 revolution = 360°

And 1 minute = 60 sec

So 33 revolution/minute $\frac{33\times 360}{60}=198^{\circ}/sec$

So angle record by vinyl $=198\times 0.23=45.54^{\circ}$ , which is nearly equal to 46

So angle record by vinyl is 46°

So option (a) will be the correct answer

5 0

3 years ago

A ball is dropped from a height of 32 feet. it bounces and rebounds 80% of the height from which it was falling

Step2247 [10]

Answer:

If it falls from 32 feet, how could the distance be 29 feet? Twelve bounces later, 58 is "obviously" incorrect as well. Eliminate those two before you do anything else.

The total distance up until that the nth bounce is

Sn = (32 - 32(.8)12) / (1 - .8) = 149.004883722... = 149

Explanation:

5 0

3 years ago

Read 2 more answers

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour her average speed is____ block/hour

S_A_V [24]

Answer:

2 m/s

Explanation:

The total time = 1 hour

The vertical displacement = 1 - 1

Vertical displacement = 0

Horizontal displacement = 4 - 2

Horizontal displacement = 2

Total displacement = sqrt (2^2 - 0^2)

Displacement - 2

Average velocity is displacement/time

= 2x1

= 2 m/s

The average velocity is 2 metres per second.

7 0

3 years ago

a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)

rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

$N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)$

Now, using the given information, we have:

$mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}$

Calculating, we have:

$N_{x}=mg*Sin(\alpha)$

$N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N$

Hence, we have that the x-component of the normal force is equal to <u>1706.45 N.</u>

Have a nice day!

3 0

3 years ago