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2 years ago

11

What is the wave speed if the period is 4.0 seconds and the wavelength is 1.8 m?

1 answer:

ELEN [110]2 years ago

7 0

Answer:

0.45 m/s

Explanation:

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A sailfish swims 120 km/hr. How far will it travel in 8.0 minutes?

Rasek [7]

Answer:

16km

Explanation:

First change the minutes into hours then multiply by the distance.

(8÷60)×120=16km

5 0

3 years ago

A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds

KengaRu [80]

First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s

3 0

3 years ago

A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back

elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity= $9.8 m/s^2$

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates= $\mu mg=0.202\times 9.8\times 35$

Force by which box accelerates= $62.286 N$

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

$ma=\mu mg$

$a=\mu g=9.8\times 0.202=1.9796 m/s^2$

Hence, the truck can have maximum acceleration before the box slide= $1.9796 m/s^2$

3 0

3 years ago

180 cm3 of hot tea at 97 °C are poured into a very thin paper cup with 20 g of crushed ice at 0 °C. Calculate the final temperat

Zolol [24]

Answer : The final temperature of the mixture is $91.9^oC$

Explanation :

First we have to calculate the mass of water.

Mass = Density × Volume

Density of water = 1.00 g/mL

Mass = 1.00 g/mL × 180 cm³ = 180 g

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of hot water (liquid) = $4.18J/g^oC$

$c_2$ = specific heat of ice (solid)= $2.10J/g^oC$

$m_1$ = mass of hot water = 180 g

$m_2$ = mass of ice = 20 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of hot water = $97^oC$

$T_2$ = initial temperature of ice = $0^oC$

Now put all the given values in the above formula, we get

$(180g)\times (4.18J/g^oC)\times (T_f-97)^oC=-(20g)\times 2.10J/g^oC\times (T_f-0)^oC$

$T_f=91.9^oC$

Therefore, the final temperature of the mixture is $91.9^oC$

8 0

3 years ago

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t) = 66 t, at which time

musickatia [10]

(x=yt-2)^2(56-tea)^2x10r

5 0

2 years ago