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forsale [732]
3 years ago
7

Can somebody help me with this

Physics
1 answer:
Ivanshal [37]3 years ago
6 0

this is simple the word your looking for is 8 letters long and the definition is key to the answer


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A bullet is fired with a muzzle velocity of 1178 ft/sec from a gun aimed at an angle of 26° above the horizontal. Find the horiz
dalvyx [7]

Answer:

1058.78 ft/sec

Explanation:

Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis

The horizontal velocity of a body can be calculated as shown below.\

Vh = Vcos∅.......................... Equation 1

Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.

Given: V = 1178 ft/sec, ∅ = 26°

Substitute into equation 1

Vh = 1178cos26

Vh = 1178(0.8988)

Vh = 1058.78 ft/sec

Hence the horizontal component of the velocity = 1058.78 ft/sec

8 0
3 years ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the
meriva

Answer:

a.2.86 m/s^2

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

\mu_s=0.14

g=9.8 m/s^2

By Newton's second law

8F-f=(8M+m)a

a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}

a=2.86 m/s^2

b.By Newton's second law

T=ma+\mu_s mg

Substitute the values

T=250\times 2.86+0.14(250)(9.8)=1058 N

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0
3 years ago
Match the speed to the section that describes.
inn [45]

Answer:

a=2 ok do it and ........

4 0
3 years ago
An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit
zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

F=qE

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

q=3.0\cdot 10^{-6}C

E=-1.5\cdot 10^4 N/C (negative because inward)

So the strength of the electrostatic force is

F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0
3 years ago
Read 2 more answers
A device for exercising the upper leg muscle is shown above, together with a schematic representation of an equivalent lever sys
yuradex [85]

yo mamaaaaaaaaaaaaaaa

4 0
3 years ago
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