Ask question

Ask question

All categories

3 years ago

14

A hungry hawk was preying on a lizard who was running northwards to get away from the low-flying hawk. If the lizard can run 8m

across water in 4s, what is the lizard's velocity? (Speed: D/T)

1 answer:

Anastaziya [24]3 years ago

8 0

Answer:

2m/s²

Explanation:

velocity = displacement (distance in a specified direction /time

You might be interested in

When an average force F is exerted over a certain distance on a shopping cart of mass m, its kinetic energy increases by 12mv2.

VMariaS [17]

Answer:

A) $d=\dfrac{1}{2F}mv^2$

B) $\Delta KE'=2\times \dfrac{1}{2}mv^2$

Explanation:

Given that

Force = F

Increase in Kinetic energy = $\dfrac{1}{2}mv^2$

$\Delta KE=\dfrac{1}{2}mv^2$

we know that

Work done by all the forces =change in the kinetic energy

a)

Lets distance = d

We know work done by force F

W= F .d

F.d=ΔKE

$F.d=\dfrac{1}{2}mv^2$

$d=\dfrac{1}{2F}mv^2$

b)

If the force become twice

F' = 2 F

F'.d=ΔKE'

2 F .d = ΔKE' ( F.d =Δ KE)

2ΔKE = ΔKE'

$\Delta KE'=2\times \dfrac{1}{2}mv^2$

Therefore the final kinetic energy will become the twice if the force become twice.

8 0

4 years ago

How much heat is needed to vaporize 10.00 grams of water at 100.0°C? The latent heat of vaporization of water is 2,259 J/g

Nata [24]

Answer:

Heat of vaporization will be 22.59 j

Explanation:

We have given mass m = 10 gram

And heat of vaporization L = 2.259 J/gram

We have to find the heat required to vaporize 10 gram mass

We know that heat of vaporization is given by $Q=mL$ , here m is mass and L is latent heat of vaporization.

So heat of vaporization Q will be = 10×2.259 = 22.59 J

8 0

3 years ago

Read 2 more answers

Can some one give me a conclusion for natural resources

levacccp [35]

Answer:

The exploitation of high-value natural resources—oil, gas, minerals, and timber—has often been a key factor in triggering, escalating, or sustaining violent conflicts around the globe. Competition over renewable resources such as land and water is on the rise, and environmental degradation, population growth, and climate change are compounding the challenges. Governments are therefore under increasing pressure to sustainably manage natural resources and resolve conflicts around their ownership, management, allocation, and control.

8 0

3 years ago

(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading i

MAVERICK [17]

Answer:

The magnitude of an earthquake is 5.6.

Explanation:

The magnitude of an earthquake can be found as follows:

$M = log(\frac{I}{S})$

Where:

I: is the intensity of the earthquake = 37.25 cm

S: is the intensity of a standard earthquake = 10⁻⁴ cm

Hence, the magnitude is:

$M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6$

Therefore, the magnitude of an earthquake is 5.6.

I hope it helps you!

6 0

3 years ago

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .

5 0

3 years ago