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3 years ago

7

a 150 kg object takes a 1.5 minute travel to a 2500 straight path it begins you trip at traveling 120 miles an hour and it decel

erates to a velocity of 20 miles an hour what is it acceleration

1 answer:

crimeas [40]3 years ago

5 0

I believe it’s acceleration is -1.11 m/s^2

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Which of these actions will increase friction? Select three options. Scratching a surface to make it rougher polishing a surface

k0ka [10]

Answer:

scratching a surface to make it rougher

increasing the size of a flying object

adding extra weight to an object

Explanation:

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3 years ago

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a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$

$\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}$

Using the formulas for $V_{ox}, V_{oy}:$

$\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}$

Simplifying

$4cos\theta sin\theta=3sin^2\theta$

Dividing by $sin\theta$

$4cos\theta=3sin\theta$

Rearranging

$tan\theta=\frac{4}{3}$

$\theta=arctan\frac{4}{3}$

$\theta=53.13^o$

4 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

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3 years ago

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

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3 years ago

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0

3 years ago