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bixtya [17]
3 years ago
12

5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on

Physics
1 answer:
kari74 [83]3 years ago
8 0

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

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The height of a person is the dependent variable because it depends on the years.
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3 years ago
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
labwork [276]

Answer:

Ein: 2.75*10^-3 N/C

Explanation:

The induced electric field can be calculated by using the following path integral:

\int E_{in} dl=-\frac{\Phi_B}{dt}

Where:

dl: diferencial of circumference of the ring

circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m

ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)

The electric field is always parallel to the dl vector. Then you have:

E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)

Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:

E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}

hence, the induced electric field is 2.75*10^-3 N/C

8 0
3 years ago
If this plastic cup is heated to its melting
Oduvanchick [21]

Answer:

A. The particles will begin to move enough

that they slide past each other.

Explanation:

When the plastic cup is heated, the Kinetic energy of its particles starts increasing. As the temperature rises, the kinetic energy keeps increasing. With the increase of K.E, the particles start moving faster and faster. When the temperature finally reaches the melting point, the K.E of the molecules is enough to break the bonds and slide past each other.

5 0
3 years ago
It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s 2 . Find the magnitude of acce
shepuryov [24]

This Question is not complete

Complete Question:

a. A hawk flies in a horizontal arc of radius 11.3 m at a constant speed of 5.7 m/s. Find its centripetal acceleration.

Answer in units of m/s2

b. It continues to fly along the same horizontal arc but increases its speed at the rate of 1.34 m/s2. Find the magnitude of acceleration under

these new conditions.

Answer in units of m/s2

Answer:

a. 2.875m/s²

b. 3.172m/s²

Explanation:

a. The formula for centripetal acceleration = (speed²) ÷ radius

Centripetal acceleration = (5.7m/s)²÷ 11.3m

Centripetal acceleration = 2.875m/s²

b. Magnitude of acceleration can be calculated by finding the sum of the vectors for the both the centripetal acceleration and the increase in the speed rate.

Centripetal acceleration ( acceleration x) = 2.875m/s²

Increase in the speed rate ( acceleration n) = 1.34m/s²

Magnitude of acceleration = √a²ₓ + a²ₙ

=√( 2.875m/s²)²+ (1.34m/s²)²

= √ 10.06m/s²

= 3.172m/s²

3 0
3 years ago
Read 2 more answers
6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove
Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0
1 year ago
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