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3 years ago

7

If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average

acceleration?

1 answer:

Nina [5.8K]3 years ago

4 0

Answer:

Average accelation = -4V

Explanation:

$a=\frac{V-V0}{t}$

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

$a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V$

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a

iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0

3 years ago

Which SI units are combined to describe a force

Keith_Richards [23]

Answer:

C.

Explanation:

kilograms and m/s^2

4 0

2 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago

The acceleration of a falling rock that was initially moving 5 m/s, and had a final velocity of 11 m/s after accelerating for 3

yanalaym [24]

Answer:

2 m/s²

Explanation:

Given:

v₀ = 5 m/s

v = 11 m/s

t = 3 s

Find: a

v = at + v₀

11 m/s = a (3 s) + 5 m/s

a = 2 m/s²

3 0

3 years ago

Why does fulcrum placement matter for how a lever works?

spin [16.1K]

Answer:

The mechanical advantage of using a lever is affected by the distance between the effort and the fulcrum and by the placement of the load. ... When the fulcrum is centered between the load and the lift, the amount of effort exerted to push down on the lever equals the amount of the load being lifted on the other end.

6 0

3 years ago