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Lubov Fominskaja [6]

2 years ago

15

A car speeds up from 18.54 m/s to 29.52 m/s in 13.84 s. The acceleration of the car is:

2 answers:

valkas [14]2 years ago

5 0

Answer:

.7934 $m/s^{2}$

Explanation:

Acceleration = change in velocity / change in time

A = 10.98 $m/s$ / 13.84 $s$

A = .7934 $m/s^{2}$

raketka [301]2 years ago

3 0

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

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) determine the density of a 32.5 g metal sample that displaces 8.39 ml of water.

sweet-ann [11.9K]

Density is the ratio of a substance's mass to its volume. On the other hand, according to Archimedes' principle, the volume of water displaced is equal to the volume of the object placed on the water. Thus, the density of the metal is equal to 8.39 mL. So, the density would be

Density = 32.5 g/8.39 mL = 3.87 g/mL

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3 years ago

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2 years ago

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

3 years ago

(a) Calculate the buoyant force on a 2.00-L helium balloon.

iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

$\rho$ = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

$F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N$

The buoyant force is 0.0239364 N

Net vertical force is given by

$F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N$

The net vertical force is 0.0057879 N

6 0

3 years ago