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erastovalidia [21]
3 years ago
13

It takes a truck 3.56 seconds to slow down from 112 km/h to 87.4 km/h. What is its average acceleration?

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

1.92 m/s2

Explanation:

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A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held b
Elina [12.6K]

Answer:

0.3 m

Explanation:

Initially, the package has both gravitational potential energy and kinetic energy.  The spring has elastic energy.  After the package is brought to rest, all the energy is stored in the spring.

Initial energy = final energy

mgh + ½ mv² + ½ kx₁² = ½ kx₂²

Given:

m = 50 kg

g = 9.8 m/s²

h = 8 sin 20º m

v = 2 m/s

k = 30000 N/m

x₁ = 0.05 m

(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²

x₂ ≈ 0.314 m

So the spring is compressed 0.314 m from it's natural length.  However, we're asked to find the additional deformation from the original 50mm.

x₂ − x₁

0.314 m − 0.05 m

0.264 m

Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.

8 0
3 years ago
Sound waves can travel through which mediums?
aksik [14]

Answer:

B. Solids, liquids, and gases.

Explanation:

I have no explanation.

5 0
3 years ago
A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°c and convection heat transfer
Elodia [21]
200 degrees because I need the points
4 0
3 years ago
A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical
loris [4]

Answer:

PE=0.92414J and KE=0.28175J

Explanation:

Gravitational potential energy=mass*gravity*height

PE=mgh

Data,

M=0.046kg

H=2.05m

g=9.8m/s^2

PE=0.046kg * 9.8m/s^2 * 2.05m

PE =0.92414J

KE=1/2mv^2

M=0.046kg

V=3.5m/s

KE=[(0.046kg)*(3.5m/s)^2]\2

KE=0.28175J

3 0
3 years ago
Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo
Ne4ueva [31]

Answer:

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

Explanation:

We have to take into account the expression for the position of the fringes

dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m

(b)  more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0
3 years ago
Read 2 more answers
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