All categories

4 years ago

What is a multiplication word problem for 15 times 15

1 answer:

3 0

"A student needs to ask for help on 15 different homework assignments, each containing 15 problems. How many problems does the student need to ask for help on altogether?"

You might be interested in

the velocity of s free fall object is given by v=√2gh where v is the velocity (m/s) g is acceleration due to gravity (m/s^2), an

Ivenika [448]

<h3>Therefore 45.87 m from which the object was dropped.</h3>

Step-by-step explanation:

Given , the velocity of free fall object is given by $v = \sqrt{2gh}$ where v is the velocity (m/s) g is acceleration due to gravity (m/s²) and h is the the distance (m) fallen.

v = 30 m/s

g = 9.81 m/s²

$v = \sqrt{2gh}$

⇔v² = 2gh [ squaring both sides]

$\Leftrightarrow h =\frac{v^2}{2g}$

$\Leftrightarrow h =\frac{30^2}{2 \times 9.81}$

$\Leftrightarrow h = 45.87$

Therefore 45.87 m from which the object was dropped.

5 0

3 years ago

Two forces with magnitudes of 100 and 50 pounds act on an object at angles of 50° and 160°, respectively. Find the direction and

MatroZZZ [7]

<h2>Hello!</h2>

Since there is no additional information, we can assume that the given angles are respect to the horizontal axis (x), so

The answer is:

Magnitude: 95.14 lbs

Direction: 80° respect to the x-axis

Since there is no additional information about the given angles, we can assume that both angles are respect to the horizontal axis (x), positive direction to x-axis positive directions (right) and above x-axis.

<h2>Calculations:</h2>

100 pounds (lb) force, 50° angle:

$100lbs*cos(50)=64.28lbs\\100lbs*sin(50)=76.60lbs$

50 pounds (lb) force, 160° angle:

$50lbs*cos(160)=-46.98lbs\\50lbs*sin(160)=17.10lbs$

So,

For x-axis forces we have:

64.28lbs (positive direction)

-46.98lbs (negative direction)

Then,

$64.28lbs-46.98lbs=17.3lbs$ (positive direction)

Fox y-axis forces we have:

76.60lbs (positive direction)

17.10 lbs (positive direction)

Then,

$76.60lbs + 17.10 lbs =93.7lbs$ (positive direction)

So, we have the two components values of the acting force:

17.3 lbs for the x-axis and 93.7 lbs for the y-axis.

Therefore,

We have to calculate the angle between both resultant force components:

Let's calll it α

So,

$\alpha=tan^{-1}(\frac{y-axisValue}{x-axisValue})=tan^{-1}(\frac{93.7}{17.3})\\\alpha=79.5=80$

Hence, the angle between both resultant force components is 80 °, meaning that the resultant force direction is positive, above the x-axis.

Now we have to calculate the magnitude of the resultant force since the sum of both resultant force components is equal to the hypotenuse of the formed triangle, we have that:

$sin(\alpha)=\frac{y-axisValue}{ResultantForceMagnitude}=\frac{93.7lbs}{ResultantForceMagnitude}\\\\ResultantForceMagnitude=\frac{93.7lbs}{sin(80)}=95.14lbs$