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strojnjashka [21]

3 years ago

8

A boy is using a rope to pull a sled to the left on a horizontal surface. The rope is parallel to the surface. What are the dire

ctions of the_______. (a) tension force and the (b) friction force on the sled

1 answer:

trasher [3.6K]3 years ago

3 0

Answer:

Explanation:

Check attachment

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two 2.5 kg balls move away from each other one traveling 3 m/s to the right the other 4 m/s to the left what is the magnitude of

Juli2301 [7.4K]

Answer:

2.5 kg.m/s

Explanation:

Taking left side as positive while right side direction as negative then

Momentum, p= mv where m is the mass of the object and v is the velocity of travel

Momentum for ball moving towards right side=mv=2.5*-3=-7.5 kg.m/s

Momentum for the ball moving towards the left side=mv=2.5*4=10 kg.m/s

Total momentum=-7.5 kg.m/s+10 kg.m/s=2.5 kg.m/s

5 0

3 years ago

How long will it take to travel 200 m traveling at 10 m/s? Follow example below.

polet [3.4K]

Answer:

$variables - d = 200m \: v = 10 m {s}^{ - 1} \\ equation \: - v = \frac{d}{t} \\ 10 = \frac{200}{t} \\ cross \: multiply \\ 10t = 200 \\ \frac{10t}{10} = \frac{200}{10} \\ t = 20s$

It will take 10 seconds to travel 200m at a speed of 10m/s

Explanation:

HOPE THAT THIS IS HELPFUL.

HAVE A GREAT DAY.

4 0

2 years ago

Read 2 more answers

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0

2 years ago

An electron follows a helical path in a uniform magnetic field of magnitude 0.340 T. The pitch of the path is 6.00 µm, and the m

dedylja [7]

Answer:

The electron's speed is 34007.35 m/s

Explanation:

It is given that,

Magnetic field, B = 0.34 T

Magnetic force on the electron, $F=1.85\times 10^{-15}\ N$

The electron follows a helical path. We have to find the speed of an electron. The formula for magnetic force is given by :

$F=B\times q\times v$

q = charge on an electron, $q=1.6\times 10^{-19}\ C$

v = velocity of an electron

$v=\dfrac{F}{Bq}$

$v=\dfrac{1.85\times 10^{-15}\ N}{0.34\ T\times 1.6\times 10^{-19}\ C}$

v = 34007.35 m/s

Hence, this is the required solution.

4 0

2 years ago

Circle the letter of each expression that has four significant figures. A. 1.25 x 10^4 B. 12.51 C. 0.0125 D. 0.1255

Andreyy89

Answer:

letter B

none zero digit are significant figures

3 0

3 years ago