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anastassius [24]

2 years ago

5

A fish is swimming around the 720-meter perimeter of her pond. If she swims 10 laps in 120 minutes, what is her average speed in

m/min

2 answers:

natulia [17]2 years ago

6 0

Her average speed is 60 m/min

zavuch27 [327]2 years ago

5 0

Answer:

60

Explanation:

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During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a

Sloan [31]

Answer:

$6.67ft/s^2$

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a= $\frac{v-u}{t}{t}$

Using the formula

Average acceleration,a= $\frac{38-18}{3}ft/s^2$

Average acceleration,a= $\frac{20}{3}ft/s^2$

Average acceleration,a= $6.67ft/s^2$

Hence, the average acceleration= $6.67ft/s^2$

5 0

2 years ago

010) Identify the true statement. Group of answer choices The height of waves is determined by wind strength and fetch. Wave bas

AlekseyPX

Answer:

The height of the wave is determined by the wind strength and fetch.

Explanation:

The height of the wave is determined by the wind strength and fetch.

The more the strength and the more the fetch size the more will be the height of the wave.

Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.

7 0

2 years ago

A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved

MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

$u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s$

Now we can solve the problem by applying the suvat equation:

$v_y^2-u_y^2=2as$

where

$v_y$ is the vertical velocity when the pea hits the ceiling

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for $v_y$ ,

$v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s$

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

$v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s$

Therefore, the speed of the pea when it hits the ceiling is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s$

5 0

2 years ago

Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and

scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U = -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0

2 years ago

When boiling water, a hot plate takes an average of 8 minutes and 55 seconds to boil 100 milliliters of water. Assume the temper

alexandr1967 [171]

Answer:

90.9 seconds

Explanation:

m = Mass of liquid = Volume×Density

c = Specific heat

$\Delta T$ = Change in temperature

t = Time taken

Room temperature = 75 °F

Converting to Celsius

$(75-32)\times \frac{5}{9}=23.889\ ^{\circ}C$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 1000\times 4186\times (100-23.889)\\\Rightarrow Q=31860.0646\ J$

Power

$P=\frac{Q}{t}\\\Rightarrow P=\frac{31860.0646}{8\times 60+55}\\\Rightarrow P=59.55152\ W$

Efficiency of the plate

$\frac{59.5512}{283}\times 100=21.04282\%$

Heat required to raise the temperature of water

$Q=mc\Delta T\\\Rightarrow Q=100\times 10^{-6}\times 784\times 2150\times (56-23.889)\\\Rightarrow Q=5412.63016\ J$

$P=\frac{Q}{t}\\\Rightarrow t=\frac{Q}{P}\\\Rightarrow t=\frac{5412.63016}{0.2104282\times 283}\\\Rightarrow t=90.9\ s$

Time taken to heat the aceton is 90.9 seconds

4 0

2 years ago