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3 years ago

Explain how elastic force is used to shoot an arrow from a bow.

Physics

2 answers:

Hunter-Best [27]3 years ago

5 0

Answer:

The bow bends when the archer pulls back on the string. The bow can bend because it is made of an elastic material. ... Releasing the string thrusts the arrow forward with the elastic force of the bow. The elastic potential energy of the bow is converted to the kinetic energy of the arrow.2 Jan 2014

mash [69]3 years ago

5 0

God does it god knows man

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The greatest pull of a magnet is near its poles.<br><br> True<br> False

DIA [1.3K]

False because opposites attract. :)

6 0

3 years ago

Read 2 more answers

A car is traveling in uniform circular motion on a section of flat roadway whose radius is . The road is slippery, and the car i

Alona [7]

Answer:

The smallest radius will be four (4) times the initial radius

Explanation:

The car maintains a constant angular speed. According to Newton's Second Law F = m a

1. $F_{r}=m*A_{n}$

2. $A_{n}=\frac{v^2}{R_{p}}$

Replacing 2 in 1

3. $F_{r}=m*\frac{v^2}{R_{p}}$

Where:

Fr= Frictional force

Rp= Initial Radius

An= Centripetal Acceleration

M= Mass

V= Velocity

Also we have that:

4. $F_{r}=\mu *W=\mu*m*g$

μ= Coefficient of friction between the car and the surface

M= Mass

W= Weight

G= Gravity

r is cleared from equation 3

5. $R_{p}=m*\frac{v^2}{F_{r}}$

Replacing 4 in 5

6. $R_{p}=m*\frac{v^2}{\mu*m*g}$

Simplifying

7. $R_{p}=\frac{v^2}{\mu*g}$

Now we have a new velocity equal to twice the initial velocity, We replace it by 2v in equation 7

8. $R_{n}=\frac{(2v)^2}{\mu*g}$

Computing

9. $R_{n}=\frac{4v^2}{\mu*g}$

Replacing 5 in 9

$R_{n}=4*R_{p}$

8 0

3 years ago

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .

5 0

3 years ago

If there was a decrease in consumer savings due to an increase in consumer spending with no increase in the money released into

lana66690 [7]

The aggregate demand curve will also decrease. If supply is not high and there is no circulating income or monetary value that's happening in a particular market, then the demand of consumers will also go down. This is because the need for production is no longer necessary because there will be no consumers to purchase goods and services from the market.

6 0

3 years ago

If a spring has an elastic potential energy of 60 J and a displacement of 4 m , what is the spring constant of the spring?

enyata [817]

Answer:

7.5 N/m

Explanation:

Potential energy of a spring can be calculated using below formula

Potential energy= 1/2kx^2

potential energy = 60 J

X= displacement = 4 m

K= spring constant=?

Substitute the values we have

60= 1/2 × K × 4^2

60= 1/2 × K × 16

60= K × 8

K= 7.5 N/m

Hence, the spring constant of the spring is 7.5 N/m

5 0

3 years ago