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Delvig [45]
3 years ago
15

A 1000 kg truck moving at 2.0 m/s runs into a concrete wall. It takes 0.5 s for the truck to completely stop. What is the magnit

ude of force exerted on the truck during the collision?​
Physics
1 answer:
guajiro [1.7K]3 years ago
3 0

Answer:

Momentum is given by

p

=

m

v

. Impulse is the change of momentum,

I

=

Δ

p

and is also equal to force times time:

I

=

F

t

. Rearranging,

F

=

I

t

=

Δ

p

t

=

0

−

20

,

000

5

=

−

4000

N

.

Explanation:

Momentum before the collision is

p

=

m

v

=

2000

⋅

10

=

20

,

000

k

g

m

s

−

1

.

Assuming the truck comes to a complete halt, the momentum after the collision is

0

k

g

m

s

−

1

.

The change in momentum,

Δ

p

, is initial minus final

→

0

−

20

,

000

=

−

20

,

000

This is called the impulse:

I

=

Δ

p

. Impulse is also equal (check the units) to force times time:

I

=

F

t

.

We can rearrange this expression to make

F

the subject:

F

=

I

t

=

Δ

p

t

=

−

20

,

000

5

=

−

4000

N

The negative sign just means the force acting is in the opposite direction to the initial momentum.

(This will be the average force acting during the collision: collisions are chaotic so the force is unlikely to be constant.)

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faltersainse [42]

Answer:

 f = 7.9487 10¹³ Hz

Explanation:

The photoelectric effect was correctly explained by Einstein assuming that the radiation is composed of photons, which behave like particles.

           hf = K + Ф

It indicates the frequency and the kinetic energy, let's look for the work function

          Ф = hf - K

let's reduce the magnitudes to the SI system

          K = 0.332 eV (1.6 10⁻¹⁹ J / 1 eV) = 0.5312 10⁻⁻¹⁹ J

let's calculate

          Ф = 6.63 10⁻⁻³⁴  6.64 10¹¹ - 0.5312 10⁻¹⁹

          Ф = 4.40 10⁻²² - 0.5312 10⁻¹⁹

          Ф = 5.27 10⁻²⁰ J

for the minimum frequency that produces photoelectrons, the kinetic energy is zero

           hf = Ф  

           f = Ф / h

           f = 5.27 10⁻²⁰ / 6.63 10⁻³⁴

           f = 7.9487 10¹³ Hz

4 0
3 years ago
A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele
kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh'            

h'=\dfrac{v^2}{2g}

h'=\dfrac{(10)^2}{2\times 9.8}

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.                              

6 0
3 years ago
Name and draw the devices that can convert digital signal to analog.
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Answer:

modem, digital music players, optical communication

Explanation:

Photos of devices are attached. An example of R2R logic circuit is also attached. DAC conversion is also done via pulse code modulation

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3 years ago
A spectroscope prism separates the white light from a star into a very wide spectrum. When widely spread, black lines appear in
topjm [15]

Answer:

C. It helps scientists detect dark matter.

Explanation:

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5 0
2 years ago
Read 2 more answers
A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere
dmitriy555 [2]

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

5 0
3 years ago
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