All categories

3 years ago

When the arrow is released, how does the the force on the arrow (by the bow) compare to the force on the archer (by the bow)?

Physics

1 answer:

Jet001 [13]3 years ago

5 0

Answer:

hello some parts of your question is missing below is the complete question

A 60.0 kg archer standing on a friction-less ice shoots a 100g arrow at a speed of 80.0 m/s to the right .When the arrow is released, how does the the force on the arrow (by the bow) compare to the force on the archer (by the bow)?

Answer : The force on the arrow is larger than the force on the archer because the arrow has a higher final velocity

Explanation:

we can compare The force on the arrow to the force on the archer by using the final velocities of the archer and the arrow ,hence The force on the arrow is larger than the force on the archer because the arrow has a higher final velocity and this is according to Newton's law which states for a every action there is an equal reaction i.e the force applied on the arrow will be the opposite of the force on the archer

You might be interested in

Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.

lbvjy [14]

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

8 0

3 years ago

A compass in a magnetic field will line up __________.

Anastaziya [24]

Line up in a direction parallel to the magnetic field lines<span />

5 0

4 years ago

A stone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the dista

Gekata [30.6K]

(a) The stone moves by uniform accelerated motion, with constant acceleration $g=9.81 m/s^2$ directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
$y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2$

(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
$750-4.9 t^2 = 0$
from which we find the time t after which the stone reaches the ground:
$t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s$

(c) The velocity of the stone at time t can be written as
$v(t) = -gt$
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
$v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s$

(d) if the stone has an initial velocity of $v_0 = 6 m/s$ , then its law of motion would be
$y(t)=y_0 - v_0t - \frac{1}{2}gt^2$
and we can find the time it needs to reach the ground by requiring again y(t)=0:
$0=750 - 6t - 4.9 t^2$
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.

5 0

3 years ago

the face of a cube towards A is brightly and shiny and the face towards V is full black.State with reason the adjustments that s

MariettaO [177]

Explanation:

increase the distance of cube from black and dull substance

5 0

3 years ago

A horizontal spring has spring constant k = 360 N/m. First, compress the spring from its uncompressed length (x = 0) to x = 11.0

mixas84 [53]

Here we can say that by energy conservation principle

Elastic potential energy of spring will convert into kinetic energy of the block