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zmey [24]
3 years ago
14

Which of the following best describes the velocity of an object? *

Physics
2 answers:
mixas84 [53]3 years ago
7 0

A. 30 m/s . . . speed

B. 30 m to the East  . . . displacement

C. 30 m/s to the East  . . . <em>velocity</em>

D. 30 m . . . distance

Elena-2011 [213]3 years ago
3 0
D fam hope this helps
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According to einstein's theory of simple relativity (_E + mc(2)_). BLANK is converted into BLANK.
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Answer:

energy is converted into mass

Explanation:

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Instantaneous speed is measured
VMariaS [17]

Answer:

C. At a particular instant

Explanation:

Speed is the defined as the ratio between the distance covered by an object and the time taken:

v=\frac{d}{t}

where d is the distance and t the time.

However, there are two possible measurements of speed:

- Average speed: this is the speed measured over a non-zero time interval (for example: a car moving 100 metres in 5 seconds; its average speed is

v=\frac{100 m}{5 s}=20 m/s

- Instantaneous speed: this is the speed of an object measured at a particular instant in time, so for a time interval that tends to zero. So, in the previous example, the average speed is 20 m/s but the instantaneous speed of the car at various instants of time can be different from that value.

7 0
3 years ago
A block of mass 20 kg sits on a ramp with an angle of 31 degrees above the horizontal. Assuming no friction, how fast will the b
Artyom0805 [142]

Answer:

2.98 m/s^2

Explanation:

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4 0
3 years ago
Read 2 more answers
A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . Part A Part complete Wha
Kisachek [45]

Answer:

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

Where,m_{1} = mass of hunter

m_{2} = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

0+0=70\times v_{1}+0.042\times590

v_{1}=-\dfrac{0.042\times590}{70}

v=-0.354\ m/s

Hence, The recoil velocity is 0.354 m/s.

8 0
3 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
3 years ago
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