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3 years ago

7

You and friend run up a fight of stairs that is 30 m high. Both of you reach the top in 12 seconds. Your weight is 570 N and you

r friend weights 620 N. Which of you has more power? Show all work

1 answer:

iren [92.7K]3 years ago

4 0

P = F*d / t
P1 = 570N * 30m / 12s = 1425W
P2 = 620N * 30m /12s = 1550W
Your friend has more power

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A dog running to the right at 4 m/s sees a ball and accelerates steadily to catch it. The dog accelerates to the right at a rate

antoniya [11.8K]

Answer:

D.-4.798m/s

Explanation:

Greetings !

Given values

$u= 4ms \\ a = 0.21ms {}^{2} \\ t = 3.8sec$

Solve for V of the given expression

Firstly, recall the velocity-time equation

$v = u + at$

plug in known values to the equation

$v = (4) + (0.21)(3.8)$

solve for final velocity

$v = 4.792ms$

Hope it helps!

6 0

1 year ago

What is the momentum of an object with 5.22 kg and 1.7 m/s

BlackZzzverrR [31]

Answer:

8.874

Explanation:

You need to times 5.22 kg and 1.7 m/s to get 8.874.

8 0

3 years ago

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid

Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0

3 years ago

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

3 years ago

Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i

mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

$\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}$

$\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad$

$w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}$

The velocity at the contact point is equal to:

$v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}$

$v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}$

Matching both expressions:

$1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s$

b) The time during which the disks slip is:

$t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s$

a) The angular acceleration of each disk is

$\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)$

$\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)$

6 0

3 years ago