Hi...... ............

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f

DedPeter [7]

Answer:

d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

d₀ = $v_{b}$ t

d₀ = 0.75 40

d₀ = 30 m

Let's write the kinematic equations

Boat

x = d₀ + $v_{b}$ t

x = 0 + $v_{h}$ t

At the meeting point the coordinate is the same for both

d₀ + $v_{b}$ t = $v_{h}$ t

t ( $v_{h}$ - $v_{b}$ ) = d₀

t = d₀ / ( $v_{b}$ - $v_{h}$ )

The two go in the same direction therefore the speeds have the same sign

t = 30 / (0.95-0.775)

t = 150 s

The distance traveled by man is

d = $v_{h}$ t

d = 0.95 150

d = 142.5 m

3 0

3 years ago

An office window has dimensions 3.5 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.96 at

Nina [5.8K]

Answer:

Force=29789.55 N

Explanation:

Given data

dimension=3.5m by 2.1m

P(outside pressure)=0.96 atm

P(inside pressure)=1.00 atm

Force=?

Solution

Area=L×W

A=(3.5)×(2.1)

$A=7.35m^{2}$

As we know

F=PA

For P(pressure)

P=P(inside)-P(outside)

P=1.00-0.96

P=0.04 atm

convert atm to pascal (pa) we get

P=4053 pa

F=PA

F=(4053)×(7.35)

F=29789.55 N

8 0

3 years ago

Rodney is trying out one of Santa's new games that consists of three pieces that blow apart if the wrong key is inserted into th

ch4aika [34]

Answer:

The third piece moves at 6.36 m/s at an angle of 65° below the horizon

Explanation:

Linear Momentum

It's a physical magnitude that measures the product of the velocity by the mass of a moving object. In a system where no external forces are acting, the total momentum remains unchanged regardless of the interactions between the objects in the system.

If the velocity of an object of mass m is $\vec v$ , the linear momentum is computed by

$\displaystyle \vec{P}=m.\vec{v}$

a)

The momentum of the board before the explosion is

$\displaystyle \vec{P}_{t1}=m_t\ \vec{v}_o$

Since the board was initially at rest

$\displaystyle \vec{P}_{t1}=$

After the explosion, 3 pieces are propelled in different directions and velocities, and the total momentum is

$\displaystyle \vec{P}_{t2}=m_1\ \vec{v}_1\ +\ m_2\ \vec{v}_2+m_3\ \vec{v}_3$

The first piece of 2 kg moves at 10 m/s in a 60° direction

$\displaystyle \vec{v}_1=(10\ m/s,60^o)$

We find the components of that velocity

$\displaystyle \vec{v}_1=$

$\displaystyle \vec{v}_1=m/s$

The second piece of 1.2 kg goes at 15 m/s in a 180° direction

$\displaystyle \vec{v}_2=(15,180^o)$

Its components are computed

$\displaystyle \vec{v}_2=(15\ cos180^o,15\ sin180^o)$

$\displaystyle \vec{v}_2=(-15,0)\ m/s$

The total momentum becomes

$\displaystyle P_{t2}=2+1.2+m_3\ \vec{v}_3$

Operating

$\displaystyle P_{t2}=++m_3\ \vec{v}_3$

Knowing the total momentum equals the initial momentum

$\displaystyle P_{t2}=+m_3\ \vec{v}_3=0$

Rearranging

$\displaystyle m_3\ \vec{v}_3=$

Calculating

$\displaystyle m_3\ \vec{v}_3=$

This is the momentum of the third piece

b)

From the above equation, we solve for $\vec v_3$ :

$\displaystyle \vec{v}_3=\frac{1}{3}$

$\displaystyle \vec{v}_3=m/s$

The magnitude of the velocity is

$\displaystyle \vec{v}_3|=\sqrt{2.67^2+(-5.77)^2}=6.36$

And the angle is

$\displaystyle tan\theta =\frac{-5.77}{2.67}=-2.161$

$\displaystyle \theta =-65.17^o$

The third piece moves at 6.36 m/s at an angle of 65° below the horizon

5 0

3 years ago

What type of air mass would form over the northern atlantic ocean?

Feliz [49]

Answer:

Maritime polar air masses form over the northern Atlantic ocean

Explanation:

Maritime polar (mP) air masses are cool, moist, and also unstable. Some maritime polar air masses arise as continental polar air masses over Asia and move westward over the Pacific, gathering warmth and moisture from the ocean. .

7 0

3 years ago

Which model correctly represents the product or products of the chemical reaction?

fgiga [73]

Answer:

you can go to quizlet in really helps.

Explanation:

7 0

3 years ago