All categories

4 years ago

What is the definition of a vector?

1 answer:

8 0

A quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.

for ex: A vector quantity has magnitude and direction.

You might be interested in

[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r

blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b) it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity $V_{f}$ for an object traveling distance h taking the initial vertical component of velocity as $V_{i}$ the kinematics equation is written as

$V_{f}^{2}=V_{i}^{2}+2ah$ where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

$V_{f}^{2}=V_{i}^{2}+2gh$

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

$V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}$ = 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

$V_{f}^{2}=V_{i}^{2}+2gh$

Since we have final velocity of 25 m/s

$V_{i}^{2}=2gh-V_{f}^{2}$

$V_{i}=\sqrt{(V_{f}^{2}-2gh)}$

$V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}$ = 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0

3 years ago

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin

Dovator [93]

Answer:

AFter 3.5 s, the wagon is moving at: $4.62\,\,\frac{m}{s}$

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

$F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}$

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

$v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}$

7 0

3 years ago

Please help 25 points!

cricket20 [7]

<h3>2 Answers:</h3>

a) Velocity is a vector quantity

e) Velocity is a speed with direction

=================================================

Explanation:

If we know the velocity of an object, then we know how fast it's going (speed) and where it's going (direction). It is a vector because the direction of the vector determines the direction, and the length of the vector (aka magnitude) determines the speed. So in a sense we've built in two facts of data into one visual.

An example of velocity: 10 miles per hour north. Here we have the speed of 10 mph and the direction north.

-------------------

Extra info:

Choice B contradicts choice A, so we can cross choice B off the list.
Choice C is false because speed is a scalar, or single quantity, and not a vector. As mentioned earlier, speed is a part of velocity, but they aren't the same exact thing.
Choice D is false because the velocity does not account for net force. We don't have any force information built into the velocity.

8 0

4 years ago

Read 2 more answers

A metallic sphere of radius 5 cm is charged such that the potential of its surface is 100 V (relative to infinity). Draw the plo

8090 [49]

Answer:

Attached below

Explanation:

The plot showing the potential as a function of distance from the center of the sphere is attached below

note : The Potential inside the sphere will remain constant as potential remains constant on surface and it start to decrease as 1/r