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dolphi86 [110]
2 years ago
5

A car advertisement states that a certain car can accelerate from rest to

Physics
2 answers:
kaheart [24]2 years ago
6 0

Answer:

a = 2.22 [m/s^2]

Explanation:

First we have to convert from kilometers per hour to meters per second

40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]

We have to use the following kinematics equation:

v_{f}= v_{i}+(a*t)

where:

Vf = final velocity = 11.11 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 5 [s]

The initial speed is taken as zero, as the car starts from zero.

11.11 = 0 + (a*5)

a = 2.22 [m/s^2]

Aliun [14]2 years ago
5 0

Refer to the attachment.

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A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

Explanation:

Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
2 years ago
1. A hydraulic lift is to be used to lift a truck masses 5000 kg. If the diameter of the
zheka24 [161]

Answer:

a)   F = 4.9 10⁴ N,  b)   F₁ = 122.5 N

Explanation:

To solve this problem we use that the pressure is transmitted throughout the entire fluid, being the same for the same height

1) pressure is defined by the relation

           P = F / A

to lift the weight of the truck the force of the piston must be equal to the weight of the truck

          ∑F = 0

          F-W = 0

          F = W = mg

          F = 5000 9.8

          F = 4.9 10⁴ N

the area of ​​the pisto is

          A = pi r²

          A = pi d² / 4

          A = pi 1 ^ 2/4

          A = 0.7854 m²

pressure is

          P = 4.9 104 / 0.7854

          P = 3.85 104 Pa

2) Let's find a point with the same height on the two pistons, the pressure is the same

          \frac{F_1}{A_1} = \frac{F_2}{A_2}

where subscript 1 is for the small piston and subscript 2 is for the large piston

          F₁ = \frac{A_1}{A_2} \ F_2

the force applied must be equal to the weight of the truck

          F₁ = ( \frac{d_1}{d_2} )^2\  m g

          F₁ = (0.05 / 1) ² 5000 9.8

          F₁ = 122.5 N

7 0
2 years ago
What is the density of a iphone with a mass of 200g and a volume of 40cm3
Leviafan [203]
Answer: 5 gm/cc

Explanation:

200 gm/40 cc
= 5 gm/cc
8 0
2 years ago
Can you answer the question
Dominik [7]
Can u show the whole question plz
3 0
3 years ago
A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn
77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0
2 years ago
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