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2 years ago

Any help is appreciated

Physics

1 answer:

sdas [7]2 years ago

8 0

image distance,di=10 cm

object distance,do=20cm

magnification, m=di/do

=10/20

=0.5

since the image is virtual, magnification is negative.

therefore m=-0.5

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The velocity of a projectile at launch has a horizontal component vh and a vertical component vv. When the projectile is at the

Usimov [2.4K]

Answer:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

Explanation:

The motion of a projectile consists of two independent motions:

- Along the horizontal direction, there are no forces acting on the projectile (if we neglect air resistance), therefore the horizontal acceleration is zero and the horizontal component of the velocity, vh, is constant

- Along the vertical direction, there is only one force acting on the projectile: the force of gravity, downward, which produces a constant downward acceleration of $g=9.8 m/s^2$ . As a consequence, the vertical component of the velocity changes according to

$v_v(t) = v_v-gt$

where vv is the initial vertical velocity and t the time. According to this equation, the vertical component of the velocity decreases first, then becomes zero at the point of maximum height, then becomes negative (= changes direction and points downward)

So, in summary, at the highest point of the trajectory we have:

- horizontal component of the velocity: $v_h$ (because it is constant)

- vertical component of the velocity: 0

- vertical component of the acceleration: $-g=-9.8 m/s^2$ (downward)

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2 years ago

Will Give Brainliest and 25 Points

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2. As the horn moves away, the sound waves become less frequent, causing the pitch to get lower.

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2 years ago

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Explanation:

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3 years ago

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Answer:

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Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

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Then, replacing the values on the equation and solving for $v_f$ , we get:

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$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

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2 years ago