All categories

2 years ago

A speedboat is moving at a rate of 45 km per hour travels a distance of 27 km. How long did it take to go to 27 km.

Physics

1 answer:

8090 [49]2 years ago

8 0

Answer:

I think it is 1.67

Explanation:

I multiply the 45 and 27 cause it says rate of which means to multiply.

You might be interested in

The combination of all forces that act on an object are (2 points)

qaws [65]

Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

5 0

3 years ago

Help would be greatly appreciated:) thank you! a pendulum clock is brought to mars. How does the bob move on Mars as compared to

Alborosie

It runs slower <span>as gravity is lower so acceleration due to gravity is smaller</span>

8 0

2 years ago

A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m

Rom4ik [11]

Answer:

2.5m

Explanation:

Torque is defined as the rotational effect of a force on a body.

The torque T for the maximum shear stress is given as 0.1 Nm

Frictional torque is the torque caused by a frictional force

The frictional torque F is given as 0.04 Nm/m

The maximum length of the shaft is thus given as

L = T / F

= 0.1/0.04

L= 2.5 m

6 0

3 years ago

Please help i don't understand. will mark brainliest

soldier1979 [14.2K]

Answer:

marblebrainiest plz\c cmarble

Explanation:

8 0

2 years ago

Read 2 more answers

A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of

zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

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

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}$

$\dfrac{1}{f}=\dfrac{347}{5852}$

$f=16.86\ cm$

We need to calculate the focal length

Using formula of magnification

$m= \dfrac{-v}{u}$

Put the value into the formula

$2=\dfrac{v}{u}$

$v = -2u$

Using formula of for focal length

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

$\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}$

$\dfrac{1}{16.86}=\dfrac{1}{2u}$

$2u=16.86$

$u=\dfrac{16.86}{2}$

$u=8.43\ cm$

Hence, The focal length is 16.86 cm and the distance of the man if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0

3 years ago