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2 years ago

15

Suppose a bird takes off from a tree and flies in a straight line. It reaches a speed of 1o miles per second. What is the change

in the bird's velocity?

1 answer:

lora16 [44]2 years ago

7 0

Answer:

10miles/second

Explanation:

Change in velocity of the bird is expressed as the difference between the final velocity and initial velocity of the body.

Change in velocity = Final velocity - initial velocity

Since the bird takes off from the tree, the initial velocity of the bird = 0miles/sec

Final velocity = 10miles/secs

Change in velocity = 10-0

Change in velocity = 10miles/second

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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0

2 years ago

Three arrows are shot horizontally. They have left the bow and are traveling parallel to the ground. Air resistance is negligibl

timurjin [86]

Answer:

F₁ = F₂ = F₃ = 0 N

Explanation:

given,

Arrow 1 mass = 80 g speed = 10 m/s

Arrow 2 mass = 80 g speed = 9 m/s

Arrow 3 mass = 90 g speed = 9 m/s

Horizontal Force:- F₁ , F₂ and F₃

There is no air resistance.

If Air resistance is zero then the horizontal acceleration of the arrow also equal to zero.

We know,

According to newton's second law

F = m a

If Acceleration is equal to zero

Then Force is also equal to zero.

Hence, F₁ = F₂ = F₃ = 0 N

4 0

2 years ago

A student uses an electronic force sensor to study how much force the student’s finger can apply to a specific location. The stu

melisa1 [442]

Answer:

B. Trial 2

Explanation:

Trial 2, because the student’s finger applied the largest force to the sensor.

Because the trial 2 student finger applied to largest force.

7 0

2 years ago

Read 2 more answers

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

Kay [80]

Explanation:

It is given that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Side of the square, L = 0.02 m

Number of turns, N = 10000

Uniform magnetic field, B = 1.5 T

Speed, v = 100 m/s

An emf is induced in the coil which is given by :

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d is the gap between the two wires connected to the ends of the coil and still get a spark. So,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Hence, this is the required solution.

6 0

2 years ago

The cheetah is considered the fastest running animal in the world. Cheetahs can accelerate to a speed of 21.7 m/s in 2.50 s and

viktelen [127]

Answer:

1) 64.2 mi/h

2) 3.31 seconds

3) 47.5 m

4) 5.26 seconds

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 0 m/s

v = Final velocity = 21.7 m/s

s = Displacement

a = Acceleration

1) Top speed = 28.7 m/s

1 mile = 1609.344 m

$1\ m=\frac{1}{1609.344}\ miles$

1 hour = 60×60 seconds

$1\ s=\frac{1}{3600}\ hours$

$28.7\ m/s=\frac{\frac{28.7}{1609.344}}{\frac{1}{3600}}=64.2\ mi/h$

Top speed of the cheetah is 64.2 mi/h

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow t=\frac{21.7-0}{2.5}\\\Rightarrow a=8.68\ m/s^2$

Acceleration of the cheetah is 8.68 m/s²

2)

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{28.7-0}{8.68}\\\Rightarrow t=3.31\ s$

It takes a cheetah 3.31 seconds to reach its top speed.

3)

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{28.7^2-0^2}{2\times 8.68}\\\Rightarrow s=47.5\ m$

It travels 47.5 m in that time

4) When s = 120 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 120=0\times t+\frac{1}{2}\times 8.68\times t^2\\\Rightarrow t=\sqrt{\frac{120\times 2}{8.68}}\\\Rightarrow t=5.26\ s$

The time it takes the cheetah to reach a rabbit is 120 m is 5.26 seconds

8 0

2 years ago