Two objects are interacting, but stay stationary. Which best describes what is happening to the action and react

A rock hits a window and stops in 0.15 seconds. The net force on the rock is 58N during the collision. What is the magnitude of

nlexa [21]

Answer:

The change in momentum is $\Delta p = 0.7 \ kg\cdot m \cdot s^{-1}$

Explanation:

From the question we are told that

The time taken for the stone to stop is $\Delta t = 0.15 \ seconds$

The net force on the rock is $F = 58 \ N$

The impulse of the rock can be mathematically represented as

$I = F * \Delta t$

Substituting values

$I = 58 * 0.15$

$I = 0.7\ kg * m * s^{-1}$

Now impulse is defined as the rate at which momentum change

Hence the change in momentum $\Delta p$ of the rock is equal to the impulse of the rock

So

$\Delta p = I = 0.7 \ kg\cdot m \cdot s^{-1}$

7 0

4 years ago

How much force is required to move a electron through an electric field with strength of 1.375 x 10^19 N/C?

Luda [366]

Answer:

The magnitude of the force required to move the electron through the given field is 2.203 N

Explanation:

Given;

The field strength of the electron, E = 1.375 x 10¹⁹ N/C

charge of electron, q = 1.602 x 10⁻¹⁹ C

The magnitude of the force required to move the electron through the given field is calculated as follows;

F = Eq

F = (1.375 x 10¹⁹ N/C) (1.602 x 10⁻¹⁹ C)

F = 2.203 N

Therefore, the magnitude of the force required to move the electron through the given field is 2.203 N

4 0

3 years ago

A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

jarptica [38.1K]

-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.

   Speed = (original speed) + (gravity x falling time)

-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is

   = (0)    + (9.8 m/s² x 3 sec)

   Velocity = 29.4 m/s downward .

6 0

3 years ago

A police officer is parked by the side of the road, when a speeding car travelling at 50 mi/hrpasses. The police car immediately

Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

speed of car $V_{c}$ = 50 mi/hr = 22.352 m/s

acceleration of police car = 10 mi/hr = 4.47 m/s²

$V_{f}$ = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

so

$V_{f}$ = $V_{i}$ + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁ = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁ will be

$d_{c}$ = $V_{c}$ × t₁

we substitute

$d_{c}$ = 22.352 × 7

$d_{c}$ = 156.46 m

now distance between polive car and speeding car

Δd = $d_{c}$ - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( $V_{f}$ - $V_{c}$ )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = $V_{f}$ × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b) how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

6 0

3 years ago

A 1000.kg car moving at north at 100. km/hr brakes to a stop in 50. m within a uniform acceleration. what are magnitude and dire

Marianna [84]

Answer:

Force=7.71KN in the opposite direction

Explanation:

Given the mass of the car(M)=1000Kg

The initial speed of the car(u)=100km/hr

we know that 1km/hr=5/18 m/sec

100km/hr=100*5/18 m/sec=27.77m/sec

the distance travelled before it stops (s)=50m

let the acceleration of the car be a

The final velocity of the car is 0.

we know that $v^{2}-u^{2}=2as$

$0^{2}-27.77^{2}=2(-a)(50)$

a=7.712m/ $sec^{2}$

We know that F=Ma

F=1000* 7.712=7712N=7.71KN

6 0

3 years ago