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Alekssandra [29.7K]

3 years ago

5

A delivery man is told to go 60 miles per hour north to make his delivery on time. These instructions grab the delivery man's( 2

points)
A. speed
B. velocity
C. acceleration
D. magnitude

2 answers:

timama [110]3 years ago

4 0

Answer:

The answer is B, velocity.

Explanation:

Just took the quiz. Hope this helps! Please count as braniliest.

GarryVolchara [31]3 years ago

4 0

Answer:

The answer is velocity.

Explanation:

Remember velocity is the speed of something in a given direction.

and beside i just took the quiz ♡

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A pendulum is suspended from the cusp of a cycloid cut in rigid support. The path described by the pendulum bob is cycloidal and

oksano4ka [1.4K]

Answer:

Verified that he oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude.

Explanation:

Starting from the first principle for the derivation and to prove that the oscillations are exactly isosynchronous with frequency ω0 = p g/l, independent of the amplitude. The mathematical manipulations was applied, trigonometric identities was also applied.The steps and explanation are shown in the attachment.

5 0

4 years ago

The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r

ollegr [7]

Answer:

The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

Explanation:

Given that,

Mass = 1.70 kg

Position vector $r=(6.00\hat{i}+5.70 t \hat{j})$

We need to calculate the angular velocity

The velocity is the rate of change of the position of the particle.

$v = \dfrac{dr}{dt}$

$v=\dfrac{d}{dt}(6.00\hat{i}+5.70 t \hat{j})$

$v=5.70\hat{j}$

We need to calculate the angular momentum of the particle

Using formula of angular momentum

$L=r\cdot p$

Where, p = mv

Put the value of p into the formula

$L=m(r\times v)$

Substitute the value into the formula

$L=1.70(6.00\hat{i}+5.70 t \hat{j}\times5.70\hat{j})$

$L=1.70\times34.2$

$L=58.14\ kgm^2/s$

Hence, The angular momentum of the particle is 58.14 kg m²/s along positive z- axis and is independent of time .

7 0

3 years ago

A raindrop of mass 0.5 * 10^-4 kg is falling verctically under the influence of gravity. The air drag on the raindrop is fdrag =

Elina [12.6K]

Answer:

The displacement of the air drop after 3 second is 18.27 m.

Explanation:

Mass of the rain drop = m = $0.5\times 10^{-4} kg$

Weight of the rain drop = W

Duration of time = t = 3 seconds

$W=m\times g$

Drag force on rain drop = $D=0.2\times 10^{-5} v^2$

$W=0.5\times 10^{-4} kg\time 10 m/s^2=0.5\times 10^{-3} N$

Motion of the rain drop:

$F=m\times a$

Net force on the rain drop , F= W - D

$W-D=m\times a$

$0.5\times 10^{-3} N-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times a$

$0.5\times 10^{-3} kg m/s^2-0.2\times 10^{-5} v^2=0.5\times 10^{-4} kg\times \frac{v}{t}$

$0.006v^2+0.05v-1.5=0$

v = 12.18 m/s

Initial velocity of the rain drop = u = 0 (since, it is starting from rest)

v=u+at (First equation of motion)

$12.18 m/s=0m/s+a\times 3 s$

$a=4.06 m/s^2$

$s=ut+\frac{1}{2}at^2$ (second equation of motion)

$s=0\times 3s+\frac{1}{2}\times 4.06m/s^2\times (3 s)^2$

s = 18.27 m

The displacement of the air drop after 3 second is 18.27 m.

6 0

3 years ago

The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident f

shtirl [24]

Answer:

17.85°

Explanation:

To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:

$n_1sin\theta_1=n_2sin\theta_2$

n1: index of refraction of Sophia's cornea = 1.387

n2: index of refraction of aqueous fluid = 1.36

θ1: angle to normal in the first medium = 17.5°

θ2: angle to normal in the second medium

You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:

$\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°$

hence, the angle to normal in the aqueous medium is 17.85°

7 0

3 years ago

What distance does a plate with an average speed of 1.95 cm/year move in 1000 years? help pls

AURORKA [14]

Answer:

1950cm

Explanation:

You should multiply 1.95 cm by 1000. This will result in the answer.

6 0

3 years ago