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g100num [7]
3 years ago
12

A ball rolls down an incline with an acceleration of 10 cm/s^2. If it starts with an initial velocity of 0 cm/s and has a velocy

of 50 cm/s when it reaches the bottom of the ramp, how long is the ramp?
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0
Given a = 10 cm/s²
          u = 0 cm/s
          v = 50 cm/s
we know that 
         v²=u²+2aS
        2500=2×10×S
        2500÷20 = S
        S= 125 cm
The ramp is 125 cm
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the velocity of a body of mass 60kg reaches 15m/s from 0m/s in 12 second. calculate the kinetic energy and power of the body.​
larisa [96]

Answer:Answer

0

Brainly User

Answer:

u = 15m/s

mass =60 kg

v =5 m/s

t = 12 sec

E k= 1/2 ×m v square

= 1/2 (15-5)×(15-5)×60

=1/2 × 10 ×10×60

30 × 100

3000J

power = energy / time

= 3000/ 12

= 250 watt

Explanation:

3 0
3 years ago
The left face of a biconvex lens has a radius of curvature of magnitude 11.7 cm, and the right face has a radius of curvature of
SIZIF [17.4K]

Answer:

Explanation:

R1 = + 11.7 cm

R2 = - 18.7 cm

n = 1.52

(a) Use lens maker formula

\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{11.7}}+\frac{1}{18.7} \right )

\frac{1}{f}=\frac{0.52\times 30.4}{218.79}

f = 13.84 cm

(b)

R1 = + 18.7 cm

R2 = - 11.7 cm

n = 1.52

(a) Use lens maker formula

\frac{1}{f}=\left ( n-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )

\frac{1}{f}=\left ( 1.52-1 \right )\left ( \frac{1}{18.7}}+\frac{1}{11.7} \right )

\frac{1}{f}=\frac{0.52\times 30.4}{218.79}

f = 13.84 cm

5 0
3 years ago
Atoms with the same atomic number but different atomic mass are called
Alexxx [7]
<span>Atoms with the same atomic number but different atomic mass are called:

<span>Isotopes</span>
</span>
4 0
2 years ago
Read 2 more answers
In what phase change do molecules completely overcome the intermolecular
Anika [276]
(B.) I know it’s not that hard or easy so don’t be made if wrong
8 0
2 years ago
Two balloons (m = 0.021 kg) are separated by a distance of d = 16 m. They are released from rest and observed to have an instant
evablogger [386]

(a) 2.56\cdot 10^{-5} C

According to Newton's second law, the force experienced by each balloon is given by:

F = ma

where

m = 0.021 kg is the mass

a = 1.1 m/s^2 is the acceleration

Substituting, we found:

F=(0.021)(1.1)=0.0231 N

The electrostatic force between the two balloons can be also written as

F=k\frac{Q^2}{r^2}

where

k is the Coulomb's constant

Q is the charge on each balloon

r = 16 m is their separation

Since we know the value of F, we can find Q, the magnitude of the charge on each balloon:

Q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(0.0231)(16)^2}{9\cdot 10^9}}=2.56\cdot 10^{-5} C

(b) 1.6\cdot 10^{14} electrons

The magnitude of the charge of one electron is

e=1.6\cdot 10^{-19}C

While the magnitude of the charge on one balloon is

Q=2.56\cdot 10^{-5} C

This charge can be written as

Q=Ne

where N is the number of electrons that are responsible for this charge. Solving for N, we find:

N=\frac{Q}{e}=\frac{2.56\cdot 10^{-5}}{1.6\cdot 10^{-19}}=1.6\cdot 10^{14}

5 0
2 years ago
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