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2 years ago

a horse moves a sleigh 1.00 kilometer by applying a horizontal 2,000 newton on its harness for 45 minutes.

1 answer:

5 0

Work:

1 kilometer = 1000 meters
45 × 60 = 2700

W = F × D
W = 2,000 N × 1,000 m
W = 2,000,000 J

P = W ÷ t
P = 2,000,000 J ÷ 2,700 s
P = 741 watts

Answer:

741 watts of horse power.

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A 20 newton force moves an object southeast. What two properties of force have been described?

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Answer:

The answer is C

Explanation:

Force is a vector quantity so it has both magnitude and direction. The question shows 20N act as magnitude and Southeast act as direction.

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2 years ago

Gravity is ____________________ to the distance between and difference in mass between two objects

olga55 [171]

Gravity is proportional to its mass and distance between it and another object

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The atomic number of an element is found by?

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2 years ago

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A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0

3 years ago

A car move with uniform acceleration along a straight line pqr .Its speed at p and r are 5m/s and 25m/s respectively if pq:qr is

topjm [15]

Answer:

Answer: Option d.

Explanation:

Accelerated Motion

When an object changes its spped in the same amounts in the same times, the acceleration is constant and its value is

$\displaystyle a=\frac{v_f-v_o}{t}$

Where $v_f, v_o, t$ are the final speed, initial speed, and time taken to change them, respectively

From the above equation we can know

$v_f=v_o+at$

The distance traveled is computed as

$\displaystyle x=v_ot+\frac{at^2}{2}$

The question talks about a car moving in a straight line with constant acceleration. It goes through the points p,q,r such as

$v_p=5\ m/s, v_r=25\ m/s$

The ratio of the distances traveled in each segment is

$\displaystyle \frac{x_1}{x_2}=\frac{1}{2}$

being $x_1$ the distance from p to q and $x_2$ the distance from q to r

It means that

$x_2=2x_1$

From the equation for speed

$v_q=v_p+at_1\ \ \ ..........[1]$

$v_r=v_q+at_2\ \ \ ..........[2]$

Replacing [1] into [2]

$v_r=v_p+at_1+at_2$

$v_r=v_p+a(t_1+t_2)$

Solving for a

$\displaystyle a=\frac{v_r-v_p}{t_1+t_2}\ \ \ .........[3]$

We now write the equation for both distances .

$\displaystyle x_1=v_pt_1+\frac{at_1^2}{2}$

$\displaystyle x_2=v_qt_2+\frac{at_2^2}{2}$

Using [1] again

$\displaystyle x_2=(v_p+at_1)t_2+\frac{at_2^2}{2}$

Since

$x_2=2x_1$

We have

$\displaystyle (v_p+at_1)t_2+\frac{at_2^2}{2}=2\left (v_pt_1+\frac{at_1^2}{2}\right )$

Operating

$\displaystyle v_pt_2+at_1t_2+\frac{at_2^2}{2}=2v_pt_1+at_1^2$

Rearranging

$\displaystyle v_pt_2-2v_pt_1=at_1^2-at_1t_2-\frac{at_2^2}{2}$

Factoring both sides

$\displaystyle v_p(t_2-2t_1)=\frac{a}{2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing the equation [3] for a :

$\displaystyle 2v_p(t_2-2t_1)=\frac{v_r-v_p}{t_1+t_2}\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Replacing $v_p=5,\ v_r=25,\ v_r-v_p=20$ , and operating the denominator

$\displaystyle 10(t_2-2t_1)\left (t_1+t_2 \right )=20\left (2t_1^2-2t_1t_2-t_2^2 \right )$

Operating and simplifying, we get a second-degree equation

$\displaystyle t_2^2+t_1t_2-2t_1^2=0$

Factoring

$(t_2-t_1)(t_2+2t_1)=0$

The only positive and valid answer is

$t_2=t_1$

Or equivalently

$\displaystyle \frac{t_1}{t_2}=1$

The option d. is correct

6 0

2 years ago