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3 years ago

A tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?

Physics

1 answer:

aliya0001 [1]3 years ago

5 0

Answer: 2.4 km/hr

Explanation:

Distance = 600m

Time= 15 minutes = 15 x 60 second/minute = 900 seconds

Speed = $\frac{Distance}{Time}$ = $\frac{600}{900}$ = $\frac{2}{3}$ m/sec

⇒ $\frac{2}{3}$ x $\frac{18}{5}$ = 2.4 km/hr (1 m/sec = 3.6 km/hr)

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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2

Nookie1986 [14]

<h2>Answers:</h2>

According to Newton's Law of Gravitation, the Gravity Force is:

$F=\frac{GMm}{{r}^{2}}$ (1)

This expression can also be written as:

$F=GMm{r}^{-2}$ (2)

If we derive this force $F$ respect to the distance $r$ between the two masses:

$\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr$ (3)

Taking into account $GMm$ are constants:

$\frac{dF}{dr}dFdr=-2GMm{r}^{-3}$ (4)

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies. </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction. </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit $N/km$ :

$\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}$ (5)

We have a force that decreases with a rate 1 $\frac{dF_{1}}{dr}dFdr=4N/km$ when $r=20000km$ :

$4N/km=-2\frac{GMm}{{(20000km)}^{3}}$ (6)

Isolating $-2GMm$ :

$-2GMm=(4N/km)({(20000km)}^{3})$ (7)

In addition, we have another force that decreases with a rate 2 $\frac{dF_{2}}{dr}dFdr=X$ when $r=10000km$ :

$XN/km=-2\frac{GMm}{{(10000km)}^{3}}$ (8)

Isolating $-2GMm$ :

$-2GMm=X({(10000km)}^{3})$ (9)

Making (7)=(9):

$(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}$ (10)

Then isolating $X$ :

$X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}$

Solving and taking into account the units, we finally have:

$X=-32N/km$ >>>>This is how fast this force changes when $r=10000 km$

7 0

4 years ago

Read 2 more answers

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is

kotykmax [81]

323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

Learn more about Tension here brainly.com/question/24994188

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2 years ago

Write the y-equation for a wave traveling in the negative x-direction with wavelength 50 cm, speed 4.0 m/s, and amplitude 5.0 cm

Zanzabum

Answer:

$y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

Explanation:

given,

wavelength, λ = 50 cm

speed, v = 4 m/s

Amplitude, A = 5 cm

general equation of the wave along x- axis

$y = A sin(2\pi(\dfrac{x}{\lambda})\pm \nu t)$

sign is positive when wave is traveling in negative direction

now,

$\nu = \dfrac{v}{\lambda}$

$\nu = \dfrac{4}{0.5}$

$\nu = 8\ s^{-1}$

inserting all the values

$y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

Hence, the y-equation of wave is equal to $y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

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3 years ago

A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3 /s. determine the minimum power that must be suppl

ELEN [110]

Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate

The minimum power required (at 100% efficiency) is
$\frac{1}{2} (8 \, \frac{m}{s} )^{2}*(9 \, \frac{m^{3}}{s} )*(1.18 \, \frac{kg}{m^{3}}) = 339.84 \, W$

The actual power will be higher because 100% efficiency is not possible.

Answer: 339.8 W (nearest tenth)

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4 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

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4 years ago

A tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?​

A tiger covers a distance of 600 m in 15 minutes what is a speed of a tiger?