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goldfiish [28.3K]

2 years ago

14

How much time will it take for a coyote to travel 48 meters across the field to get to the unsuspecting rabbit eating grass in t

he field? He is travelling at 4 m/s

1 answer:

kkurt [141]2 years ago

4 0

Formula for time
t=d/s
so…
t= 48m/4m/s
the two ms cancel each other out and ur left with s

t=12s

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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di

dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

Στ = 0

τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

+ W d₁ - w_cat d₂ = 0

d₂ = W / w d₁

d₂ = M /m d₁

d₂ = 5.00 /2.9 0.850

d₂ = 1.466 m

6 0

3 years ago

Change the following data into scientific notation.<br><br> 5 000 000 km

RoseWind [281]

To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000.

8 0

3 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

A tennis ball, 0.314 kg, is accelerated at a rate of 164 m/s2 when hit by a professional tennis player. What force does the play

SVEN [57.7K]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

   = (0.314 kg) x (164 m/s²)

   = 51.5 newtons

   (about 11.6 pounds) .

Notice that the ball is only accelerating while it's in contact with the racket.
The instant the ball loses contact with the racket, it stops accelerating, and
sails off in a straight line at whatever speed it had when it left the strings.

4 0

3 years ago

1. You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a

svetoff [14.1K]

Answer:

Explanation:

1 ) Magnetic field due to a circular coil carrying current

= μ₀I / 2r

I is current , r is radius of the wire , μ₀ = 4π x 10⁻⁷

= 4π x 10⁻⁷ x 15 / (2 x 3.5 x 10⁻²)

= 26.9 x 10⁻⁵ T

2 )

Negative z direction .

The direction of magnetic field due to a circular coil having current is known

with the help of screw rule or right hand thumb rule.

3 )

If we decrease the radius the magnetic field will:__increase _____.

A. Increase.

Magnetic field due to a circular coil carrying current

B = μ₀I / 2 r

Here r is radius of the coil . If radius decreases magnetic field increases.

So magnetic field will increase.

4 0

2 years ago