Which formula can be used to find the angle of the resultant vector?

What is the term used for sediments building up at the mouth of the river? (Hint: think New Orleans)

meriva

I believe it’s called Alluvium! It’s where the river mouth is build up of gravel,sand,silt, and clay!!

4 0

3 years ago

A cannon is fired horizontally at 243 m/s off of a 62 meter tall, shear vertical cliff. How far in meters from the base of the c

Alekssandra [29.7K]

Answer:

865.08 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 243 m/s

Height (h) of the cliff = 62 m

Horizontal distance (s) =?

Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:

Height (h) of the cliff = 62 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

62 = ½ × 9.8 × t²

62 = 4.9 × t²

Divide both side by 4.9

t² = 62/4.9

Take the square root of both side.

t = √(62/4.9)

t = 3.56 s

Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:

Initial velocity (u) = 243 m/s

Time (t) = 3.56 s

Horizontal distance (s) =?

s = ut

s = 243 × 3.56 s

s = 865.08 m

Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.

6 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos

lakkis [162]

Answer:

a) x = 4.33 m , b) w = 2 rad / s , f = 0.318 Hz , c) a = - 17.31 cm / s²,

d) T = 3.15 s, e) A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

x = 5 cos (π / 6)

x = 4.33 m

remember angles are in radians

b) The general form of the equation is

x = A cos (w t + Ф)

when comparing the two equations

w = 2 rad / s

angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 2 / 2pi

f = 0.318 Hz

c) the acceleration is defined by

a == d²x / dt²

a = - A w² cos (wt + Ф)

for t = 0 , we substitute

a = - 5,0 2² cos (π / 6)

a = - 17.31 cm / s²

d) El period is

T = 1/f

T= 1/0.318

T = 3.15 s

e) the amplitude

A = 5.0 cm

3 0

3 years ago

what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?

Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0

3 years ago

Read 2 more answers

An unknown radioactive sample is observed to decrease in activity by a factor of two in a one hour period. What is its half-life

elena55 [62]

Answer:

The half-life is $t_{1/2} = 1.005 h$

Explanation:

Using the decay equation we have:

$A=A_{0}e^{-\lambda t}$

Where:

λ is the decay constant
A(0) the initial activity
A is the activity at time t

We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that $A = \frac{A_{0}}{2}$

$\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}$

$0.5=e^{-\lambda*1 h}$

Taking the natural logarithm on each side we have:

$ln(0.5)=-\lambda$

$\lambda=0.69 h^{-1}$

Now, the relationship between the decay constant λ and the half-life t(1/2) is:

$\lambda = \frac{ln(2)}{t_{1/2}}$

$t_{1/2} = \frac{ln(2)}{\lambda}$

$t_{1/2} = \frac{ln(2)}{0.69}$

$t_{1/2} = 1.005 h$

I hope it helps you!

6 0

3 years ago