All categories

3 years ago

How will the motion of the arrow change after it leaves the bow?

1 answer:

4 0

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

You might be interested in

A rocket blasts off vertically from rest on the launch pad with a constant upward acceleration of 2.30 m/s2. At 20.0 s after bla

lesya [120]

Answer:

Explanation:

v = u +at

u = 0

a = 2.3 m /s²

t = 20 s

v = 2.3 x 20

= 46 m /s

Distance covered under acceleration of 2.3 m/s²

s = ut + 1/2 at²

= 0 + .5 x 2.3 x 20²

= 460 m

After that it moves under free fall ie g acts on it downwards .

v² = u² - 2gh , h is height moved by it under free fall

0 = 46² - 2 x 9.8 h

h = 107.96 m

Total height attained

= 460 + 107.96

= 567.96 m

b ) At its highest point ,it stops so its velocity = 0

c ) rocket's acceleration at its highest point = g = 9.8 downwards .

At highest point , it is undergoing free fall so its acceleration = g

6 0

3 years ago

What is the momentum of a 5 kg object that has a velocity of 1. 2 m/s? 3. 8 kg â€˘ m/s 4. 2 kg â€˘ m/s 6. 0 kg â€˘ m/s 6. 2 kg â

nevsk [136]

The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

Momentum of a substance is the product of its mass and velocity. That is;

Momentum (p) = mass (m) × velocity (v)

According to this question, an object has a mass of 5kg and velocity of 1.2m/s. The momentum is calculated thus:

Momentum = 5kg × 1.2m/s

Momentum = 6kgm/s.

Therefore, the momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

Learn more about momentum at: brainly.com/question/250648?referrer=searchResults

7 0

2 years ago

Read 2 more answers

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

ArbitrLikvidat [17]

Answer:

L = 499 m

Explanation:

If we assume that the speed of sound is constant, that travels along a straight line, and that the echo is instantaneous, we can find the total distance travelled by the sound, as follows, just applying the definition of average velocity:

$\Delta x = v_{s} * t = 343 m/s* 2.91 s = 998 m$

If we assume that the time needed to reach to the cliff, is the same used for the return travel, the length of the lake will be exactly half of the total distance calculated:

$l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m$

The length of the lake is 499 m.

8 0

3 years ago

Individuals who suffer from a mental disorder are most likely to commit violent crimes. true or false

Damm [24]

That is true. Some people with mental disorders envision the world differently. Some people think that people are out to get them, they are in another world, etc. They will do anything to feel safe in their mental state.

I hope this helps!
~cupcake

7 0

3 years ago

Read 2 more answers