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3 years ago

A car enters a tunnel at 24 m/s and accelerates steadily at 2.0 m/s2. At what speed does it leave the tunnel, 8.0 seconds later?

1 answer:

4 0

Given:

V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
V2 = final velocity = ?

For linear-motion problems with those given terms, the following formula is used:

V2 = V1 + as

Substituting the given values:

V2 = 24 + 2(8)
V2 = 24 + 16
V2 = 40 m/s

Therefore the car will have a speed of 40 m/s as it leaves the tunnel.

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If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to

mamaluj [8]

Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.

7 0

3 years ago

Read 2 more answers

An intergalactic rock star bangs his drum every 1.50 s. A person on earth measures that the time between beats is 2.70 s. How fa

sineoko [7]

Answer:

v = 83.1 % of speed of light

Explanation:

given,

T_e is the earth time = 2.7 s

T_s is the ship time = 1.5 s

we know,

$T_s = T_e \times \gamma$

where c is the speed of light

v is the speed of the rock star moving

$T_s = T_e\times \sqrt{1-\dfrac{v^2}{c^2}}$

$1.5= 2.7\times \sqrt{1-\dfrac{v^2}{c^2}}$

$\sqrt{1-\dfrac{v^2}{c^2}} =0.556$

squaring both side

$1-\dfrac{v^2}{c^2}=0.3086$

$v^2=0.6914c^2$

v = 0.831 c

v = 83.1 % of speed of light

7 0

3 years ago

During a hurricane, the atmospheric pressure inside a house may blow off the roof because of the reduced pressure outside. If ai

m_a_m_a [10]

Answer:

817.5 Pa

Explanation:

From Bernoulli's equation, considering thst there is no height difference then

P1+½d(v1)²=P2+½d(v2)²

P1-P2=½d(v2²-v1²)

∆P=½d(v2²-v1²)

Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure

Given the speed at roof top as 128 km/h, we convert it to m/s as follows

128*1000/3600=35.555555555555=35.56 m/s

Velocity at the bottom of roof is 0 m/s

Density is given as 1.293 kg/m³

∆P=½*1.293*(35.56²-0)=817.5 Pa

5 0

3 years ago

A person will feel weightless whenever?

gavmur [86]

When your'e in space you feel weightless blc there's no grsvity in space.

5 0

3 years ago

Read 2 more answers

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago