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A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent

Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0

3 years ago

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

An object with mass 100 kg moved in outer space. When it was at location <8, -30, -4> its speed was 5.5 m/s. A single cons

Alenkasestr [34]

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

sin φ = z / r

Cos φ = r_p / r

z = r sin φ

r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

r = √126

r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

φ = sin⁻¹ z / r

φ = sin⁻¹ ( $\frac{-7+4}{11.23}$ )

φ = 15.5º

Let's find the projection of the position vector (r_p)

r_p = r cos φ

r_p = 11.23 cos 15.5

r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

cos θ = x / r_p

θ = cos⁻¹ x / r_p

θ = cos⁻¹ ( $\frac{14-8}{10.82}$ )

θ = 56.3º

taking the angles we can decompose the initial velocity.

sin φ = v_z / v₀

cos φ = v_p / v₀

v_z = v₀ sin φ

v_z = 5.5 sin 15.5 = 1.47 m / z

v_p = vo cos φ

v_p = 5.5 cos 15.5 = 5.30 m / s

cos θ = vₓ / v_p

sin θ = v_y / v_p

vₓ = v_p cos θ

v_y = v_p sin θ

vₓ = 5.30 cos 56.3 = 2.94 m / s

v_y = 5.30 sin 56.3 = 4.41 m / s

we already have the components of the initial velocity

v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

Fₓx = m aₓ₁

aₓ₁ = Fₓ / m

aₓ₁ = 220/100

aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

let's calculate

vₓ₁² = 2.94² + 2 2.20 (14-8)

vₓ₁ = √35.04

vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

aₓ₂ = Fₓ₂ / m

aₓ₂ = 100/100

aₓ₂ = 1 m / s²

final speed

v_{xf}² = vₓ₁² + 2 aₓ₂ (x_f- x₁)

v_{xf}² = 5.92² + 2 1 (17-14)

v_{xf} =√41.05

v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

a_{y1} = F_y / m

a_{y1} = 460/100

a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

v_{y1} = √102.25

v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

a_{y2} = F_{y2} / m

a_{y2} = 260/100

a_{y2} = 2.60 m / s2

final speed

v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

v_{yf} = √ 71.01

v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

first interval, relate (a_{z1})

a_{z1} = F_{z1} / m

a_{z1} = -200/100

a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

v_{z1} = √14.16

v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})