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3 years ago

Clem's credit score is 733, while Ingrid's credit score is 688. how much more would Ingrid have to pay per month than Clem?

2 answers:

5 0

The answer is 64. you subtract how much they actually pay.
Clems score is 733 which is 924 on the chart. Ingrids score is 688 which allows her to pay 860 per month ..
924-860= 64

Verizon [17]3 years ago

4 0

Answer:

$64

Explanation:

Apex

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An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

$W=0$

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3 years ago

Let the four horizontal compass directions north, east, south, and west be represented by units vectors n^, e^, s^, and w^, resp

Marrrta [24]

Answer:

c = e > b = d > a

Explanation:

Given vectors are all unit vectors, therefore they have a magnitude of 1

<h3>Let a, b be two vectors and magnitude of cross product of these two vectors is (magnitude of a) × (magnitude of b) × (sine of angle between these two vectors)</h3>

As all are unit vectors their magnitude is 1 and therefore in this case the cross product between any two vectors depends on the sine of angle between those two vectors

In option a as both the vectors are same, the angle between them will be zero and sin0° will also be 0

In option b angle between those two vectors is 135° and sin135° is 1 ÷ √2

In option c angle between those two vectors is 90° and sin90° is 1

In option d angle between those two vectors is 45° and sin45° is 1 ÷ √2

In option e angle between those two vectors is 90° and sin90° is 1

So by comparison of magnitudes of cross products in each option, the order will be c = e > b = d > a

5 0

3 years ago

31. If you threw a baseball straight out at 45 m/s from a height of 1.5 meters (A) how long would it be in the air? B) How far o

coldgirl [10]

Answer:

A) t = 0.55 s

B) x = 24.8 m

Explanation:

A) We can find the time at which the ball will be in the air using the following equation:

$y_{f} = y_{0} + v_{0y}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ is the final height= 0

$y_{0}$ is the initial height= 1.5 m

$v_{0y}$ is the component of the initial speed in the vertical direction = 0 m/s

t: is the time =?

g: is the gravity = 9.81 m/s²

$0 = 1.5 m - \frac{1}{2}9.81 m/s^{2}t^{2}$

By solving the above equation for t we have:

$t = \sqrt{\frac{2*1.5 m}{9.81 m/s^{2}}} = 0.55 s$

Hence, the ball will stay 0.55 seconds in the air.

B) We can find the distance traveled by the ball as follows:

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

Where:

a: is the acceleration in the horizontal direction = 0

$x_{f}$ is the final position =?

$x_{0}$ is the initial position = 0

$v_{0x}$ is the component of the initial speed in the horizontal direction = 45 m/s

$x_{f} = x_{0} + v_{0x}t + \frac{1}{2}at^{2}$

$x_{f} = 0 + 45 m/s*0.55 s + 0 = 24.8 m$

Therefore, the ball will travel 24.8 meters.

I hope it helps you!

3 0

2 years ago

When we say Mohit has Blue Eyes which part of the ice is begin referred to US blue

oksian1 [2.3K]

Answer:

The iris

Explanation:

6 0

2 years ago

Read 2 more answers

USE K U E S

nata0808 [166]

Explanation:

Problem 1.

Initial speed of the runner, u = 0

Acceleration of the runner, $a=4.2\ m/s^2$

Time taken, t = 100 s

Let v is the speed of the runner now. Using the first equation of kinematics as :

$v=u+at$

$v=at$

$v=4.2\ m/s^2\times 100\ s$

v = 420 m/s

Problem 2.

Initial speed of the plane, u = 0

Distance covered, d = 300 m

Time taken, t = 25 s

Using the equation of kinematics as :

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}at^2$

$a=\dfrac{2d}{t^2}$

$a=\dfrac{2\times 300\ m}{(25\ s)^2}$

$a=0.96\ m/s^2$

Problem 3.

A ball free falls from the top of the roof for 5 seconds. Let it will fall at a distance of d. It is given by :

$d=ut+\dfrac{1}{2}gt^2$

$d=\dfrac{1}{2}\times 9.8\times (5)^2$

d = 122.5 meters

Let v is the final speed at the end of 5 seconds. Again using first equation of kinematics as :

$v=u+gt$

$v=9.8\ m/s^2\times 5\ s$

v = 49 m/s

Hence, this is the required solution.

3 0

3 years ago