All categories

3 years ago

A truck slows from a velocity of 25 m/s to a stop in 70 m. What was the truck’s acceleration

Physics

1 answer:

igomit [66]3 years ago

4 0

v^2-u^2=2 x a x d

25^2-0^2=2 x a x 70

625-0=140 x a

625=140a

a=625/140

a=4.46 m/s^2

im not very sure but i think this is how you do this

You might be interested in

A physicist is calibrating a spectrometer that uses a diffraction grating to separate light in order of increasing wavelength (λ

cluponka [151]

Answer:

Explanation:

Given

$N=3680 cm^{-1}$

therefore slit spacing $d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm$

since $d\sin \theta =n\lambda$

for $n=1$

$d\sin \theta =\lambda$

Now,at $\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm$

at $\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm$

at $\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm$

5 0

3 years ago

Dimensional formula of current density

kozerog [31]

Ampere per meter squared A/m $x^{2}$

6 0

3 years ago

Read 2 more answers

Two iron weights, one twice the mass of the other, are dropped from the top of a building. Compared with the lighter weight, the

Whitepunk [10]

The twice as heavy weight will hit the ground with more force, or impact.

4 0

3 years ago

Read 2 more answers

Explain how tidal forces are causing Earth to slow down.

dybincka [34]

Answer: The tidal forces exerted by the moon are directly associated with the earth's rotation. Due to the strong gravitational pull of the moon, the tidal bulging appears on both the sides on earth and these are region of high tide, and there is gradual rise and fall of sea level.

Because of these tidal effect, the earth is able to rotate only once in each of the orbital period.

6 0

3 years ago

La ecuación de la posición de una esferita está dada por: r(t)=(2.Cos(πt) i-3.Sen(πt) j) (m) ¿Cuál es la velocidad de la esferit

katovenus [111]

Answer:

v = (-4.44 i^ + 6.66 j^ ) m/s, a_average =( 0 i^ -2π j^) m/s²

Explanation:

The expression left corresponds to an oscillatory movement (MAS), the speed is defined by

v = dr / dt

the function of position

r = 2 cos πt i^ + 3 sin πt j^

let us note that it is a movement in two dimensions

let's perform the derivative

v = -2π sin πt i^ + 3π cos πt j^

we evaluate this expression for t = 0.25 s, remember that the angle is in radians

v = -2π sin (π 0.25) i^ + 3π cos (π 0.25) j^

v = (-4.44 i^ + 6.66 j^ ) m/s

To calculate the mean acceleration we use the expression

a = ( $v_{f} - v_{o}$ ) / Δt

indicates that the time is the first 3 s

we look for the initial velocity t = 0 s

v₀ = 0 i ^ + 3π j ^

we look for the fine velocity, t = 3 s

v_f = - 2π sin (π 3) + 3π cos (π 3) j ^

v_f = 0 i ^ - 3π j ^

we calculate the average acceleration

Δt = (3 -0) = 3 s

a_average = (0-0) / 3 i ^ + (-3π - 3π) / 3

a_average = (0 i ^ -2π j ^ ) m/s²

6 0

3 years ago