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3 years ago

I need help with part D

Physics

1 answer:

natka813 [3]3 years ago

8 0

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

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Two bumper cars move in a straight line with the following equations of motion: x1 = -4.0 m + (1.1 m/s )t x2 = 8.8 m + (-2.9 m/s

lawyer [7]

Answer:

3.2 seconds

Explanation:

When the cars collide, they have the same position.

x₁ = x₂

-4 + 1.1t = 8.8 − 2.9t

4t = 12.8

t = 3.2

5 0

2 years ago

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

5 0

3 years ago

How much force is required to move a sled 5 meters if a person uses 60 j of work? 300N 12N 65N 30N

maksim [4K]

The answer is b 12N because
We know that
W = F × d × c o s(θ)
assuming theta=0 we then solve and have
F=W/d
substitute known values to get:
F=60J/5m=12N

4 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

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3 years ago

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Anni [7]

Answer:

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2 years ago

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