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2 years ago

13

How to the magnitude of the applied force of a suspended string which applies a horizontal force to make an angle to the vertica

l

1 answer:

nataly862011 [7]2 years ago

8 0

Answer:

Vector have magnitude and direction

Explanation:

Use the component X and component y to find the magnitude and direction

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What would a series circuit be used for?

igor_vitrenko [27]

Answer:

C

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0

2 years ago

psychology chapter 7, principles of learning; I am the process whereby animals are taught a complicated response by being reward

jeyben [28]

<span>Answer: Burrhus Frederic Skinner's Operant Conditioning.

</span><span>B.F. Skinner believed that to understand behavior, in the best way, is to look at the root causes or reasons of an action and its outcomes.
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Skinner proposes the Law of Effect-Reinforcement. Here,he differentiated the positively reinforced behavior or the strengthened behavior, the negatively reinforced behavior (removal of the unpleasant experience), and weakened behavior because of punishment.
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In positive reinforcement, behavior is strengthened through providing an outcome, an effect that an individual finds rewarding. Negative reinforcement also strengthens behavior because the unpleasant experience was removed. Punishment on the other hand is an opposite to reinforcement. Instead of increasing the response, it eliminates it or weakens it.
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8 0

3 years ago

a 74.9 kg person sits at rest on an icy pond holding a physics book. he throws the physics book west at 8.25 m/s and he recoils

kifflom [539]

Answer:

1.95 kg

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (74.9) (-0.215) + m (8.25)

m = 1.95

3 0

2 years ago

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a

vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

3 0

3 years ago

Read 2 more answers

Explain how mirrors can produce images that are larger or smaller than life size, as well as upright or inverted

galina1969 [7]

Answer:

1) When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

2) When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

3) When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror

Explanation:

The position of an object in front of a concave mirror of radius of curvature, R, determines the size and orientation of the image of the object as illustrated in the mirror equation

$\dfrac{1}{f}=\dfrac{1}{d_{o}} + \dfrac{1}{d_{i}}$

$Magnification, \, m = \dfrac{h_{i}}{h_{o}} = -\dfrac{d_{i}}{d_{o}}$

Where:

f = Focal length of the mirror = R/2

$d_{i}$ = Image distance from the mirror

$d_{o}$ = Object distance from the mirror

$h_{i}$ = Image height

$h_{o}$ = Object height

$d_{o}$ is positive for an object placed in front of the mirror and negative for an object placed behind the mirror

$d_{i}$ is positive for an image formed in front of the mirror and negative for an image formed behind the mirror

m is positive when the orientation of the image and the object is the same

m is negative when the orientation of the image and the object is inverted

f and R are positive in the situation where the center of curvature is located in front of the mirror (concave mirrors) and f and R are negative in the situation where the center of curvature is located behind the mirror (convex mirrors)

∴ When $d_{o}$ < $d_{i}$ (hence $d_{o}$ < f ) and they are both in front of the mirror (positive), the image will be larger and inverted

When $d_{o}$ > $d_{i}$ (and $d_{o}$ < f ) such that they are both positive (in front of the mirror), the image will be smaller and inverted

When the image is behind the mirror, for convex mirrors and the object is in front the image will be uptight. The magnification of the image will be the ratio of the image distance to the object distance from the mirror.

5 0

3 years ago