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Nikitich [7]
4 years ago
13

A 1-kilogram object is thrown horizontally and a 2-kilogram object is dropped vertically at

Physics
1 answer:
Semenov [28]4 years ago
4 0
Their vertical position will be the same for both at any instant. They will both hit the ground at the same time.
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If a light ray strikes a flat mirror at an angle of 44° from horizontal the reflected ray will be​
ladessa [460]

Answer: The reflected ray will be 44 degrees.

Explanation: The Law of Reflection states the angle of incidence ( the incoming light ray) equals the angle of reflection ( the light ray being reflected) Therefore, the angle of reflection will be 44 degrees.

Hope that helped!

3 0
3 years ago
If a force of 5 n to left and 9 n to right act on a object what is the net force
emmainna [20.7K]
4 n to the right is the net force
7 0
3 years ago
List at least 3 units of speed circle the unit of speed which is the SI unit
Stells [14]

Answer: SI Unit is meters per second(m/s)

Explanation:

The three units of speed are:

1. meters per second(m/s)

2. kilometers per hour(km/hr)

3. miles per hour(mph)

SI Unit is meters per second(m/s)

6 0
3 years ago
9. A car driver brakes gently. Her car slows down front --
sleet_krkn [62]

Answer:

9) This is a case of deceleration

10)-0.8 ms-2

b) acceleration is the change in velocity with time

11)

a) 100 ms-1

b) 100 seconds

12) 10ms-1

13) more information is needed to answer the question

14) - 0.4 ms^-2

15) 0.8 ms^-2

Explanation:

The deceleration is;

v-u/t

v= final velocity

u= initial velocity

t= time taken

20-60/50 =- 40/50= -0.8 ms-2

11)

Since it starts from rest, u=0 hence

v= u + at

v= 10 ×10

v= 100 ms-1

b)

v= u + at but u=0

1000 = 10 t

t= 1000/10

t= 100 seconds

12) since the sprinter must have started from rest, u= 0

v= u + at

v= 5 × 2

v= 10ms-1

14)

v- u/t

10 - 20/ 25

10/25

=- 0.4 ms^-2

15)

a=v-u/t

From rest, u=0

8 - 0/10

a= 8/10

a= 0.8 ms^-2

7 0
4 years ago
The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire
kakasveta [241]

Answer:

F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

Vector (OA) = 6i + 0j + 0k \\Vector (OB) = 0i + 3j + 2k \\Vector (OC) = 0i - 2j + 3k

Next step is to find unit vectors u_{AB} ,u_{AC}, u_{AD}, u_{AE} as follows:

u_{AB} = \frac{vector(AB)}{magnitude(AB)} \\= \frac{OB - OA}{magnitude({vector(OB - OA))} }\\=\frac{-6i +3j+2k}{\sqrt{6^2 + 3^2+2^2} } \\\\=-0.857 i +0.429j+0.286k\\\\u_{AC} = \frac{vector(AC)}{magnitude(AC)} \\= \frac{OC - OA}{magnitude({vector(OC - OA))} }\\=\frac{-6i -2j-3k}{\sqrt{6^2 + 2^2+3^2} } \\\\=-0.857 i -0.286j+0.429k\\\\u_{AD} = +1i\\u_{AC} = -1k

Using the diagram we find the corresponding vectors Forces:

F_{AB} = F_{AB} i + F_{AB}j +F_{AB}k\\F_{AC} = F_{AC} i + F_{AC}j +F_{AC}k\\F_{AD} = F_{AD} i + F_{AD}j +F_{AD}k\\W = -160 k

Equation of Equilibrium:

Sum of forces = 0\\F_{AB}. u_{AB} + F_{AC}.u_{AC} + F_{AD}.u_{AD} + W = 0\\(-0.857F_{AB}i + 0.429F_{AB}j +0.286F_{AB}k) + (-0.857F_{AC}i - 0.286F_{AC}j +0.429F_{AC}k) + (+1F_{AD} i)  + (-160k) = 0

Comparing i , j and k components as follows:

-0.857F_{AB} -0.857F_{AC}  +1F_{AD}  = 0\\+ 0.429F_{AB} - 0.286F_{AC} = 0\\+0.286F_{AB} +0.429F_{AC}  =  160

Solving Above equation simultaneously we get:

F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004

3 0
4 years ago
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